= Solution
The polynomial of $\alpha$ satisfies the <Eisenstein criterion> at $p$. Let $\mathfrak p$ be any <prime ideal> of $\mathcal O_K$ above $p$, normalize its ideal valuation by $v_{\mathfrak p}(\mathfrak p)=1$, and write $e=v_{\mathfrak p}(p)$ for its <ramification index of a prime ideal>. Put $t=v_{\mathfrak p}(\alpha)$. In
$$
\alpha^n+a_{n-1}\alpha^{n-1}+\cdots+a_0=0,
$$
the constant term has valuation $e$, while every other $a_j\alpha^j$ has valuation at least $e+jt$. The <non-Archimedean valuation> property then forces
$$
nt=e.
$$
Indeed, either strict inequality would leave a unique term of least valuation in the equation. Since $t$ is a positive integer and $e\le n$, it follows that $t=1$ and $e=n$.
The fundamental inequality $\sum_{\mathfrak p\mid p}e_{\mathfrak p}f_{\mathfrak p}\le n$ now shows that $\mathfrak p$ is the unique prime above $p$ and has residue degree one. At this prime,
$$
v_{\mathfrak p}(p)=n,\qquad v_{\mathfrak p}(\alpha)=1,
$$
while $(p,\alpha)$ has no factor away from $p$. Unique <prime ideal factorization> therefore gives
$$
\boxed{P=(p,\alpha)=\mathfrak p,\qquad P^n=(p)}.
$$
Finally $v_P(\alpha)=1$, so
$$
\boxed{\alpha\notin P^2}.
$$
This is <total ramification from an Eisenstein polynomial>.
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