= Solution
Evaluation at $\alpha$ gives the ring isomorphism
$$
\mathcal O_L\cong\mathbb Z[X]/(f),
$$
and reducing modulo $p$ gives
$$
\mathcal O_L/p\mathcal O_L
\cong\mathbb F_p[X]/(\overline f).
$$
For each $i$,
$$
\mathcal O_L/P_i
\cong\mathbb F_p[X]/(\overline g_i).
$$
This quotient is a field because $\overline g_i$ is irreducible. Hence $P_i$ is a nonzero <prime ideal>; it is nonzero because it contains $p$. Its <ideal norm> is
$$
\boxed{N(P_i)=|\mathcal O_L/P_i|
=p^{\deg\overline g_i}.}
$$
The ideals are distinct. Indeed, if $P_i=P_j$ with $i\ne j$, then both $\overline g_i$ and $\overline g_j$ would vanish in the same quotient. Since distinct monic irreducibles are coprime, a <Bezout identity> between them would make $1$ vanish there, a contradiction.
Because the chosen lifts satisfy
$$
f(X)=\prod_{i=1}^r g_i(X)^{e_i}+pH(X)
$$
for some $H\in\mathbb Z[X]$, evaluation at $\alpha$ gives
$$
\prod_i g_i(\alpha)^{e_i}\in p\mathcal O_L.
$$
Every generator of the product ideal $\prod_iP_i^{e_i}$ either contains a factor $p$ or is the displayed product, so
$$
\prod_iP_i^{e_i}\subseteq p\mathcal O_L.
$$
On the other hand, multiplicativity of ideal norms and $\sum_i e_i\deg\overline g_i=\deg f=[L:\mathbb Q]$ give
$$
N\!\left(\prod_iP_i^{e_i}\right)
=p^{\sum_i e_i\deg\overline g_i}
=p^{[L:\mathbb Q]}
=N(p\mathcal O_L).
$$
Two finite-index ideals with one contained in the other and equal index are equal. Thus the <Dedekind factorization theorem> is proved in this monogenic case:
$$
\boxed{p\mathcal O_L=P_1^{e_1}\cdots P_r^{e_r}.}
$$
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