It might be useful to abstract notation a bit: we are considering some simple extension $K=\mathbb Q_p(\alpha)$ of $\mathbb Q_p$ where $\alpha\in\mathcal O_K$ has minimal polynomial $f(x)\in\mathbb Z_p[x]$ which is irreducible mod $p$. Let's call $\kappa$ the residue field of $K$. Since $\alpha\in\mathcal O_K$ we have a residue $\overline\alpha\in\kappa$. Therefore we have a tower in characteristic $p$
$$\mathbb F_p\subseteq\mathbb F_p(\overline\alpha)\subseteq\kappa.$$
But $\overline\alpha$ is a root of $\overline f(x)\in\mathbb F_p[x]$, which is irreducible, so you can conclude that
$$[\mathbb F_p(\overline\alpha):\mathbb F_p]=\deg(\overline f)=\deg(f)=[K:\mathbb Q_p]$$
from which you deduce $[\kappa:\mathbb F_p]=[K:\mathbb Q_p]$ and the extension is unramified.
Edit to address the question you're making in the comment: let's write an equation for $f(\alpha)=0$, so we give $f$ some coefficients, say $f(x)=\sum_{i=0}^n c_ix^i$ where $c_i\in\mathbb Z_p$ and $c_n=1$. Then we have an equation in $\mathcal O_K$:
$$0=f(\alpha)=\sum_{i=0}^n c_i\alpha^i.$$
Now we consider this equation modulo $\mathfrak m_K$: in $\kappa$ we have
$$0=\sum_{i=0}^n(c_i+\mathfrak m_K)(\alpha+\mathfrak m_K)^i.$$
Your question (if I'm understanding correctly) is why this actually shows that $\alpha+\mathfrak m_K$ is a root of $\overline f$, the residue of $f$ modulo $p$, since above we have coefficients mod $\mathfrak m_K$. The point is simple (but the notation is ugly): the following square commutes
$$\require{AMScd}
\begin{CD}
\mathbb Z_p @>{i}>> \mathcal O_K\\
@VVV @VVV \\
\mathbb F_p @>{\iota}>> \kappa
\end{CD}$$
and so $c_i+\mathfrak m_K=\iota(c_i+p\mathbb Z_p)$ which gives us
$$0=\sum_{i=0}^n(c_i+\mathfrak m_K)(\alpha+\mathfrak m_K)^i=\sum_{i=0}^n \iota(c_i+p\mathbb Z_p)(\alpha+\mathfrak m_K)^i=(\iota(\overline f))(\alpha+\mathfrak m_K)$$
where we write $\iota(\overline f)$ for the image of $\overline f$ under the map $\mathbb F_p[x]\to\kappa[x]$ induced by $\iota$, and we see that $\alpha+\mathfrak m_K$ is a root of $\iota(\overline f)$, then we abuse notation and say "$\alpha+\mathfrak m_K$ is a root of $\overline f$" (we are always implicitly doing this type of thing when we work with arbitrary field extensions!)
I guess the last point is why the square commutes: this is really just unraveling what it means to consider $\kappa/\mathbb F_p$ as an extension in the first place; the way we say this is an extension is by having a homomorphism $\mathbb F_p\to\kappa$, and the most concrete way to get this homomorphism is to consider the composition $\mathbb Z_p\to\mathcal O_K\to\kappa$, and notice that this map kills $p$ (since $p\in\mathfrak m_K$) and therefore it factors through a map on the quotient $\iota:\mathbb F_p=\mathbb Z_p/(p)\to\kappa$.
