Hint: The inertial degree is the degree of the field extension $\mathbb{Z}/(p) = \mathbb{F}_p \subset \mathbb{Z}[\alpha]/\mathcal{P}$.
Given what you already know, the only sensible choice of the polynomial $f$ is to take $f$ to be the minimal polynomial of $\alpha$ over $\mathbb{Q}$, which since $\alpha$ is integral, will actually have coefficients in $\mathbb{Z}$.
Now there are infinitely map primes $p$ for which $f(x) \equiv 0 (mod p)$ has a solution $r_p$ in $\mathbb{Z}$. Consider the ideal $(\alpha - r_p, p) \subset \mathbb{Z}[\alpha]$.
We would like $(\alpha - r_p, p)$ to be a prime ideal. To show this, observe that we have $\mathbb{Z}[\alpha]/(\alpha - r_p, p) \cong \mathbb{Z}[x]/(f(x), x - r_p , p) = \mathbb{Z}[x]/(x-r_p,p)$.
Why does the last equality hold (hint divide $f(x)$ by $x - r_p$ and deduce that the constant remainder is a multiple of $p$ from the fact that $f(r_p) \equiv 0 (mod \hspace{1mm} p)$)?
What is the ring $\mathbb{Z}[x]/(x - r_p , p) $ isomorphic to?
Now can you conclude that $(\alpha - r_p, p) \cap \mathbb{Z} = (p)$.
I think my solution is okay (can someone confirm this?) given that in your question $\mathcal{O}_K = \mathbb{Z}[\alpha]$. Things would have been slightly more complicated if $\mathcal{O}_K$ was just the ring of integers of $K = \mathbb{Q}(\alpha)$ for an algebraic interger $\alpha$.