$p$-adic valuation of $P_n(p)$ if $P_n$ is the $n$th Legendre polynomial

Question

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

Answer 1

This is a suggestion, in case it helps.

The two sequences satisfy the following recurrences, respectively, \begin{align} nP_n(p)&=p(2n-1)P_{n-1}(p)-(n-1)P_{n-2}(p) \\ nM_n(p)&=p(2n-1)M_{n-1}(p)-(n-1)M_{n-2}(p)\cdot(p-2)^2. \end{align}