$\text{Hom}_k(P,P)$ is isomophic to $P\otimes_k \text{Hom}_k(P,k)$ Let $P$ be a finitely generated projective right module over $k$. Then,
$$\text{Hom}_k(P,P)\cong P \otimes_k \text{Hom}_k(P,k).$$
I was able to show the congruence assuming $P$ is a vector space over $k$.
The map was $f \colon P\otimes_k \text{Hom}_k(P,k) \to \text{Hom}_k(P,P)$ such that $$f(p,g)(p_0)=p.g(p_0)$$ for $p,p_0\in P$ and $g\in \text{Hom}_k(P,k)$. Then using the vector space dimension argument would give us the isomorphism.
But I don't have much idea as how to prove it for projective modules.
 A: I will assume that $k$ is just a commutative ring with unity.
Also, I will write $P^\vee$ for $\text{Hom}_k(P,k)$, and for every $k$-module $M$ denote by $\varphi_M$ to the map $M \otimes_k P^\vee \to \text{Hom}_k(P,M)$ such that $m \otimes g \mapsto g(\_)m$ (so, $\varphi_P$ is the map that you describes).
Note that if $f \colon M_1 \to M_2$ is any $k$-module homomorphism, we have a commutative diagram
$$
\require{AMScd}
\begin{CD}
M_1 \otimes_k P^\vee @>{f \otimes 1}>> M_2 \otimes_k P^\vee \\
@VV{\varphi_{M_1}}V @VV{\varphi_{M_2}}V \\
\text{Hom}_k(P,M_1) @>>{f_*}> \text{Hom}_k(P,M_2)
\end{CD}
$$
in which $f_*$ is pre-composition with $f$.
Now, if $P$ is projective and finitely generated, $P$ is a direct summand of a free and finitely generated $k$-module $F$.
Therefore, we have maps $i \colon P \to F$ and $\pi \colon F \to P$ such that $\pi \circ i = 1$.
Thus, from the diagram
$$
\require{AMScd}
\begin{CD}
P \otimes_k P^\vee @>{i \otimes 1}>> F \otimes_k P^\vee @>{\pi \otimes 1}>> P \otimes_k P^\vee \\
@VV{\varphi_P}V @VV{\varphi_F}V @VV{\varphi_P}V \\
\text{Hom}_k(P,P) @>>{i_*}> \text{Hom}_k(P,F) @>>{\pi_*}> \text{Hom}_k(P,P)
\end{CD}
$$
we observe the following:

*

*Since $(\pi \otimes 1) \circ (i \otimes 1) = 1$ and $\pi_* \circ i_* = 1$, the two horizontal arrows of the first (resp. second) square are injective (resp. surjective).

*If $\varphi_F$ is injective, then so is $\varphi_F \circ (i \otimes 1) = i_* \circ \varphi_P$, which implies that $\varphi_P$ is injective too.

*If $\varphi_F$ is surjective, then so is $\pi_* \circ \varphi_F = \varphi_P \circ (\pi \otimes 1)$, which implies that $\varphi_P$ is surjective too.

Hence, if we prove that $\varphi_F$ is an isomorphism, we’re done.
Indeed, since $F = k^n$ for some $n \in \Bbb N$, consider the projections $p_1,\dots,p_n \colon F \to k$, and note that the diagram
$$
\require{AMScd}
\begin{CD}
F \otimes_k P^\vee @>{\varphi_F}>> \text{Hom}_k(P,F) \\
@V{(a_1,\dots,a_n) \otimes g \,\mapsto\, (a_1g,\dots,a_ng)\ }VV @VV{f \,\mapsto\, (p_1 \circ f,\dots,p_n \circ f)}V \\
(P^\vee)^n @= (P^\vee)^n
\end{CD}
$$
commutes.
Because the vertical arrows are isomorphisms, then so is $\varphi_F$.
