Is it true that $\operatorname{Hom}_R(M,N) \otimes_R P=\operatorname{Hom}_R(M ,N \otimes P)$? Let $M, N, P$ be modules over a commutative ring $R$. The above identity is true for $R$ a field:
Since the RHS $\cong$ (by passing to double dual)  $$\operatorname{Hom}(N^* \otimes P^*, M^*)\cong\operatorname{Hom}(P^*,\operatorname{Hom}(N^*,M^*)) \cong P^{**}  \otimes\operatorname{Hom}(M,N)\cong P \otimes\operatorname{Hom}(M,N).$$
In a couple examples that I worked out, it held true but the steps do not hold true if $N$ and $M$ have torsion.  Is the identity true?    

edit:  I had originally made a mistake when writing my question.  I have changed the question title to make it correct.  Originally I had written $\operatorname{Hom}_R(M,N) \otimes_R P=\operatorname{Hom}_R(M\otimes P,N)$ which as Qiaochu Yuan remarks is ludicrous (for the vector space case $RHS= P^*\otimes\operatorname{Hom}(M,N)$). Other than cosmetic changes the body of the text remain unchanged.                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                              
 A: As explained here, this does not hold in general, but it does hold if any of the following sets of hypotheses holds:


*

*$P$ is finitely presented projective,

*$M$ is finitely presented and $P$ is flat,

*$P$ is finitely presented and $M$ is projective.


As an example of the second case, $P$ can be a localization of the base ring, and you get that if $M$ is finitely presented then hom commutes with localization. 
Your argument can be made to work under the additional assumption of dualizability, which for modules is equivalent to being finitely presented projective; see for example this blog post for details. But you assume in your argument that $M, N, P$ are all dualizable, and as mentioned above you actually only need to assume that $P$ is dualizable, although this doesn't cover all cases of interest. 
A: It is true if $P$ is finitely generated projective, for instance. In this case you have $P\oplus Q=R^n$ and the statement boils down to prove
$$\def\H{\operatorname{Hom}_R}
\H(M,N)\otimes R^n\cong\H(M,N\otimes_RR^n)
$$
which is almost obvious.
If $R$ is a field, $P$ finitely generated projective means $P$ is finite dimensional.
In general you can only say there's a natural map
$$
\H(M,N)\otimes_RP\to\H(M,N\otimes_RP)
$$
given by the bilinear map
$$
\H(M,N)\times P\to\H(M,N\otimes_RP),
\qquad
(f,p)\mapsto \widetilde{(f,p)}
$$
where
$$
\widetilde{(f,p)}\colon m\mapsto f(m)\otimes p
$$
A: I am going to put Prof. Suarez-Alvarez's comment as an answer.
Right so it is not true in the case for infinite dimensional vector spaces because I need to use double duality to get the identity $Hom(V,W)=V^* \otimes W$.  In more detail $Hom(V,W) = Hom (V, Hom(W^*,R))=Hom(V \otimes W^* ,R)=(V\otimes W^*)^*=V^* \otimes W$.  I used this identity in the second isomorphism in my question.
Nor is it true when $M=\mathbb{Z}/2$, $N=\mathbb{Z}$, $P=\mathbb{Z}/2$, for LHS=$Hom(\mathbb{Z}/2,\mathbb{Z})\otimes \mathbb{Z}/2=0\otimes \mathbb{Z}/2=0$ and $RHS=Hom(\mathbb{Z}/2,\mathbb{Z}/2)=(\mathbb{Z/2})^4$.
