This answer will prove the following.
Theorem 1. Let $X$ be a Banach space. Let $p, p' \in (1, \infty )$ such that $\tfrac{1}{p} + \tfrac{1}{p'} = 1$. The following conditions are equivalent.
$({\rm a})$ The Banach space $X$ is isomorphic to $\ell_{p}$.
$({\rm b})$ There exist sequences $(x_{j})_{j\in\mathbb{N}}$ in $X$ and $(x_{j}^{*})_{j\in\mathbb{N}}$ in $X^{*}$ that satisfy the conditions $({\rm i}) {\rm -} ({\rm iv})$ below.
$({\rm i})$ For all $j,k\in\mathbb{N}$ we have
\begin{equation}
x_{j}^{*}(x_{k}) = \begin{cases}
1 & \text{if } j = k, \\
0 & \text{if } j \neq k .
\end{cases}
\end{equation}
$({\rm ii})$ For every $x\in X$ we have $(x_{j}^{*}(x))_{j\in\mathbb{N}} \in \ell_{p}$.
$({\rm iii})$ For every $x^{*} \in X^{*}$ we have $(x^{*}(x_{j}))_{j\in\mathbb{N}} \in \ell_{p'}$.
$({\rm iv})$ We have $\overline{{\rm span}\{x_{j}^{*} : j\in\mathbb{N}\} } = X^{*}$.
Before proving the theorem above we will need some lemmas. Likely all of them apart from potentially the first will be known to the reader.
Lemma 1. Let $X$ be a Banach space and $p\in [1, \infty ]$. Let $(x_{j}^{*})_{j\in\mathbb{N}}$ be a sequence in $X^{*}$ such that $(x_{j}^{*}(x))_{j\in\mathbb{N}} \in \ell_{p}$ for every $x\in X$. Define $T\colon X \to \ell_{p}$ by $Tx := (x_{j}^{*}(x))_{j\in\mathbb{N}}$. Then $T$ is a bounded linear operator.
Proof. It is straightforward to show that $T$ is linear. We will show that $T$ has a closed graph and apply the closed graph theorem. Let $(x_{k})_{k\in\mathbb{N}}$ be a sequence in $X$. Let $x\in X$ and $y\in \ell_{p}$. Assume the sequence $(x_{k})_{k\in\mathbb{N}}$ converges to $x$ in $X$ and the sequence $(Tx_{k})_{k\in\mathbb{N}}$ converges to $y$ in $\ell_{p}$. We will show $Tx = y$. Let $j\in\mathbb{N}$. Define $e_{j}^{*} \colon \ell_{p} \to \mathbb{K}$ by $e_{j}^{*}(a) := a(j)$. Then $e_{j}^{*} \in \ell_{p}^{*}$. We have
\begin{equation}
y(j) = e_{j}^{*}(y) = \lim_{k\to\infty} e_{j}^{*}(Tx_{k}) = \lim_{k\to\infty} x_{j}^{*}(x_{k}) = x_{j}^{*}(x) = (Tx)(j) .
\end{equation}
As $j\in\mathbb{N}$ was arbitrary we deduce $Tx = y$. Hence $T$ has a closed graph. It now follows from the closed graph theorem that $T$ is a bounded linear operator.
Lemma 2. Let $X$ and $Y$ be Banach spaces. Let $S$ and $T$ be bounded linear operators from $X$ into $Y$. Then the set $\{x\in X : Sx = Tx \}$ is a closed subspace of $X$.
Proof. Observe that $\{x\in X : Sx = Tx \}$ is the kernel of the bounded linear operator $S - T$ and the kernel of a bounded linear operator is a closed subspace of the domain.
Lemma 3. Let $p, p' \in (1, \infty )$ such that $\tfrac{1}{p} + \tfrac{1}{p'} = 1$. We have the following.
$({\rm i})$ Let $a^{*} \in \ell_{p}^{*}$. For each $j\in\mathbb{N}$ define $e_{j} \colon \mathbb{N} \to \mathbb{K}$ by
\begin{equation}
e_{j}(k) := \begin{cases}
1 & \text{if } j = k, \\
0 & \text{if } j \neq k .
\end{cases}
\end{equation}
Then $e_{j} \in \ell_{p}$ for all $j\in\mathbb{N}$. We have $(a^{*}(e_{j}) )_{j\in\mathbb{N}}$ with $\| (a^{*}(e_{j}) )_{j\in\mathbb{N}} \|_{\ell_{p'}} = \| a^{*} \|_{\ell_{p}^{*} }$.
$({\rm ii})$ Define $\Phi \colon \ell_{p'} \to \ell_{p}^{*}$ by
\begin{equation}
(\Phi a)(b) := \sum_{j=1}^{\infty} a(j) b(j) .
\end{equation}
Then $\Phi$ is a well-defined isometric isomorphism.
Proof. See here.
We now prove the following. Note that the assumptions are formulated slightly differently to the question.
Proof of Theorem 1. $({\rm a}) \Rightarrow ({\rm b})$. For each $j\in\mathbb{N}$ define $e_{j} \colon \mathbb{N} \to \mathbb{K}$ by
\begin{equation}
e_{j}(k) := \begin{cases}
1 & \text{if } j = k, \\
0 & \text{if } j \neq k ,
\end{cases}
\end{equation}
and $e_{j}^{*} \colon \ell_{p} \to \mathbb{K}$ by $e_{j}(a) := a(j)$. We have that $e_{j} \in \ell_{p}$ and $e_{j}^{*} \in \ell_{p}^{*}$ for every $j\in\mathbb{N}$ as well as
\begin{equation}
e_{j}^{*}(e_{k}) = \begin{cases}
1 & \text{if } j = k , \\
0 & \text{if } j \neq k
\end{cases}
\end{equation}
for all $j, k \in \mathbb{N}$. Also recall that $(a^{*}(e_{j}) )_{j\in\mathbb{N}} \in \ell_{p'}$ for every $a^{*} \in \ell_{p}^{*}$. Let $T\colon X\to \ell_{p}$ be an isomorphism. Since $T$ is surjective, for each $j\in\mathbb{N}$ we find some $x_{j}\in X$ such that $T x_{j} = e_{j}$. Also define $x_{j}^{*} := T^{*} e_{j}^{*}$ for each $j\in\mathbb{N}$. We will show that the sequences $(x_{j})_{j\in\mathbb{N}}$ in $X$ and $(x_{j}^{*})_{j\in\mathbb{N}}$ satisfy the conditions $({\rm i}) {\rm -} ({\rm iv})$ from the statement of Theorem 1.
We first show $({\rm i})$. Let $j, k\in\mathbb{N}$. We have
\begin{equation}
x_{j}^{*}(x_{k}) = (T^{*} e_{j}^{*} )(x_{k}) = e_{j}^{*} (Tx_{k}) = e_{j}^{*} (e_{k}) = \begin{cases}
1 & \text{if } j = k , \\
0 & \text{if } j \neq k .
\end{cases}
\end{equation}
This shows $({\rm i})$. We now show $({\rm ii})$. Let $x\in X$ and $N\in\mathbb{N}$. We have
\begin{equation}
\sum_{j=1}^{N} |x_{j}^{*}(x) |^{p} = \sum_{j=1}^{N} | (T^{*} e_{j}^{*})(x) |^{p} = \sum_{j=1}^{N} | e_{j}^{*}(Tx) |^{p} = \sum_{j=1}^{N} |(Tx)(j) |^{p} \leq \| Tx\|_{\ell_{p}}^{p} .
\end{equation}
As $N\in\mathbb{N}$ was arbitrary we deduce $(x_{j}^{*}(x))_{j\in\mathbb{N}} \in \ell_{p}$. This shows $({\rm ii})$. We now $({\rm iii})$. Since $T$ is invertible we see from the equations $T^{*} (T^{-1})^{*} = (T^{-1} T)^{*} = {\rm id}_{X^{*}}$ and $(T^{-1})^{*} T^{*} = (T T^{-1})^{*} = {\rm id}_{\ell_{p}^{*}}$ that $T^{*}$ is invertible with $(T^{*})^{-1} = (T^{-1})^{*}$. In particular $T^{*}$ is surjective. Let $a^{*} \in \ell_{p}^{*}$ and $N\in\mathbb{N}$. Using Lemma 3 for the last inequality of $(1)$, we have
\begin{equation}
\sum_{j=1}^{N} | (T^{*}a^{*})(x_{j}) |^{p'} = \sum_{j=1}^{N} | a^{*}(Tx_{j}) |^{p'} = \sum_{j=1}^{N} | a^{*}(e_{j}) |^{p'} \leq \| a^{*} \|_{\ell_{p}^{*} }^{p'} . \tag{1}
\end{equation}
As $N\in\mathbb{N}$ was arbitrary we deduce $(x_{j}^{*}(x))_{j\in\mathbb{N}} \in \ell_{p}$. Since $T^{*}$ is surjective this shows $({\rm iii})$.
We now show $({\rm iv})$. Since $T$ is an isomorphism we have
\begin{equation}
\ell_{p} = \overline{{\rm span} \{e_{j} : j\in\mathbb{N} \} } = \overline{{\rm span} \{Tx_{j} : j\in\mathbb{N} \} } = T(\overline{{\rm span}\{x_{j} : j\in\mathbb{N} \} } )
\end{equation}
and hence
\begin{equation}
X = T^{-1}(\ell_{p}) = T^{-1}( T(\overline{{\rm span}\{x_{j} : j\in\mathbb{N} \} } ) ) = \overline{{\rm span}\{x_{j} : j\in\mathbb{N} \} } .
\end{equation}
This shows $({\rm iv})$. Hence the sequences $(x_{j})_{j\in\mathbb{N}}$ in $X$ and $(x_{j}^{*})_{j\in\mathbb{N}}$ satisfy the conditions $({\rm i}) {\rm -} ({\rm iv})$ from the statement of Theorem 1.
$({\rm b}) \Rightarrow ({\rm a})$. Define $T\colon X\to \ell_{p}$ by $Tx := (x_{j}^{*}(x))_{j\in\mathbb{N}}$ and $S \colon X^{*} \to \ell_{p}^{*}$ by $Sx^{*} := (x^{*}(x_{j}))_{j\in\mathbb{N}}$. Note that $T$ and $S$ are well defined by $({\rm ii})$ and $({\rm iii})$ respectively. Define $J\colon X\to X^{**}$ by $(Jx)(x^{*}) := x^{*}(x)$. We first apply Lemma 1 using $X$ and the sequence $(x_{j}^{*})_{j\in\mathbb{N}}$ to see that $T$ is a bounded linear operator. We then apply Lemma 1 using $X^{*}$ and the sequence $(Jx_{j})_{j\in\mathbb{N}}$ to see that $S$ is also a bounded linear operator. Define $\Phi \colon \ell_{p'} \to \ell_{p}^{*}$ by
\begin{equation}
(\Phi a)(b) := \sum_{j=1}^{\infty} a(j) b(j) .
\end{equation}
We have that $\Phi$ is an isometric isomorphism by this result. We now outline the rest of the proof. We will show the equalities
\begin{equation}
T^{*} \circ \Phi \circ S = {\rm id}_{X^{*}} \tag{1}
\end{equation}
and
\begin{equation}
\Phi \circ S \circ T^{*} = {\rm id}_{\ell_{p}^{*}} . \tag{2}
\end{equation}
From $(1)$ and $(2)$ we will deduce that $T^{*}$ is an isomorphism and conclude from there that $T$ is an isomorphism between $X$ and $\ell_{p}$.
We first show $(1)$. Let $j\in \mathbb{N}$ and $x\in X$. Using $({\rm i})$ for the last equality, we have
\begin{equation}
\begin{aligned}
( (T^{*} \circ \Phi \circ S) x_{j}^{*} )(x) &= ( T^{*}( \Phi (Sx_{j}^{*} ) ) ) (x) \\
&= \Phi (Sx_{j}^{*}) (Tx) \\
&= \sum_{k=1}^{\infty} (Sx_{j}^{*})(k) \, (Tx)(k) \\
&= \sum_{k=1}^{\infty} x_{j}^{*}(x_{k}) \, x_{k}^{*}(x) \\
&= x_{j}^{*}(x) .
\end{aligned} \tag{3}
\end{equation}
Because $x\in X$ was arbitrary, we deduce from $(3)$ that
\begin{equation}
(T^{*} \circ \Phi \circ S) x_{j}^{*} = x_{j}^{*} . \tag{4}
\end{equation}
As $j\in\mathbb{N}$ was arbitrary, we see that $(4)$ holds for all $j\in\mathbb{N}$. Consequently, by Lemma 2 the set $\{x^{*}\in X^{*} : (T^{*} \circ \Phi \circ S) x^{*} = {\rm id}_{X^{*}} x^{*} \}$ is a closed subspace of $X^{*}$ containing $\{x_{j}^{*} : j\in\mathbb{N} \}$. It now follows from $({\rm iv})$ that $\{x^{*}\in X^{*} : (T^{*} \circ \Phi \circ S) x^{*} = {\rm id}_{X^{*}} x^{*} \} = X^{*}$. Hence $T^{*} \circ \Phi \circ S = {\rm id}_{X^{*}}$.
We now show $(2)$. For each $j\in\mathbb{N}$ define $e_{j}^{*} \colon \ell_{p} \to \mathbb{K}$ by $e_{j}^{*}(a) := a(j)$. Then $e_{j}^{*} \in \ell_{p}^{*}$ for every $j\in\mathbb{N}$. Let $j\in\mathbb{N}$ and $a\in \mathbb{N}$. We have
\begin{equation}
\begin{aligned}
( (\Phi \circ S \circ T^{*} ) e_{j}^{*} )(a) &= \Phi (S(T^{*} e_{j}^{*} ) ) (a) \\
&= \sum_{k=1}^{\infty} (S (T^{*} e_{j}^{*}) )(k) \, a(k) \\
&= \sum_{k=1}^{\infty} (T^{*} e_{j}^{*})(x_{k}) \, a(k) \\
&= \sum_{k=1}^{\infty} e_{j}^{*} (Tx_{k}) \, a(k) \\
&= \sum_{k=1}^{\infty} (Tx_{k})(j) \, a(k) \\
&= \sum_{k=1}^{\infty} x_{j}^{*}(x_{k}) \, a(k) \\
&= a(j) = e_{j}^{*}(a) .
\end{aligned} \tag{5}
\end{equation}
Because $a\in \ell_{p}$ was arbitrary, we deduce from $(5)$ that
\begin{equation}
(\Phi \circ S \circ T^{*}) e_{j}^{*} = e_{j}^{*} . \tag{6}
\end{equation}
As $j\in\mathbb{N}$ was arbitrary, we see that $(6)$ holds for all $j\in\mathbb{N}$. Consequently, by Lemma 2 the set $\{a^{*}\in \ell_{p}^{*} : (\Phi \circ S \circ T^{*}) a^{*} = {\rm id}_{\ell_{p}^{*}} a^{*} \}$ is a closed subspace of $\ell_{p}^{*}$ containing $\{e_{j}^{*} : j\in\mathbb{N} \}$. It now follows from $({\rm iv})$ that $\{a^{*}\in \ell_{p}^{*} : (\Phi \circ S \circ T^{*}) a^{*} = {\rm id}_{\ell_{p}^{*}} a^{*} \} = \ell_{p}^{*}$. Hence $\Phi \circ S \circ T^{*} = {\rm id}_{\ell_{p}^{*}}$.
We have now shown both $(1)$ and $(2)$. Combining $(1)$ and $(2)$, we deduce that $T^{*}$ has a bounded inverse given by $(T^{*})^{-1} = \Phi \circ S$. Hence $T^{*}$ is an isomorphism between $\ell_{p}^{*}$ and $X^{*}$. We can now take adjoints and use that $\ell_{p}$ is reflexive as done in the answer by mihaild, or alternatively use this result, which has a proof only relying on Hahn-Banach theorems (see here), to conclude that $T$ is an isomorphism between $X$ and $\ell_{p}$. Hence $X$ is isomorphic to $\ell_{p}$ as desired. This completes the proof.
