Here is sufficient condition for obtaining the isomorphism you are after.
Let $A$ be an arbitrary commutative ring (not supposed noetherian) and $M, N$ arbitrary modules . We have a canonical morphism
$$ M^*\otimes_A N\to Hom (M,N):\phi \otimes n \mapsto [m\mapsto\phi(m)n] \quad (\star)$$
Proposition ($\star$) is an isomorphism as soon as $M$ is finitely generated projective.
It is an isomorphism for $M=A$, then for finitely generated free modules $M=A^r$, and finally for summands of such i.e. finitely generated projectives.
Given three arbitrary modules $M,N,P$ over the commutative ring $A$ , with $M $ finitely generated projective, we have a natural isomorphism
$$Hom_A(M,N)\otimes_A P \cong Hom_A(M,N\otimes_A P)$$
Replace all $Hom$'s by $\otimes$'s and use associativity of tensor product.
a) The Corollary fails if $P$ is finitely generated but not projective.
Take for example $A=\mathbb Z, N=\mathbb Z, M=P=\mathbb Z/(2)$.
Then the left-hand side in the Corollary is $0$ and the right-hand side is $\mathbb Z/(2)$
b) As QiL pertinently comments, the Corollary also fails if $P$ is not assumed finitely generated.
Take for example $N=A$ and $M=P=$ an arbitrary non finitely generated module.
Then our canonical morphism $u: M^*\otimes M \to Hom(M,M)$ cannot be surjective.
Indeed any element $t\in M^*\otimes M$ gets sent to an endomorphism $u(t)=f:M\to M$ such that $u(M)$ is finitely generated, so that the identity $Id_M$ of $M$ will never be in the image of $u$.