In 3-adic valuation of a sum involving binomial coefficients the $p$-adic valuation of $P_n(p)$ has been obtained.

Computations indicate that $M_n(p)=\sum\binom{n}{k}^2(p-1)^k$ for odd $p$ has the same $p$-adic valuation:
$$\nu_p (P_n(p)) = \nu_p (M_n(p)).$$

Is this true for all odd $p$?

(Note that $P_n(3)=M _n(3)$).