Ok so the problem is the following: $\newcommand{\Hom}{\operatorname{Hom}}$
Given $M$ a finitely generated and projective left $R$-module, prove that $\Hom_{R}(M,R)$ seen as a right $R$-module is also projective. Here $(fr)(m):=f(rm)$ for all $m\in M$, $f \in \Hom_{R}(M,R)$ and $r \in R$
My current attempt:
Since $M$ is finitely generated and proyective, define a surjective map $\pi : R^n \to M$, where the map is defined by where it sends the basis of $M$.
Now as $M$ is proyective we have that $M \oplus \ker \pi \cong R^n$. We also know that there exist a $Z(R)$-isomorphism $\varphi : \Hom_{R}(\bigoplus_{1}^n R,R) \rightarrow \bigoplus_1^n \Hom_R(R,R)$, where $Z(R)$ is the center of the ring.
On the other hand we also have a $Z(R)$-isomorphism $$\phi: \Hom_{R}(M\oplus \ker\pi, R) \rightarrow \Hom_R(M,R) \oplus \Hom_R(\ker\pi,R).$$ We also have that $$\Hom_{R}(M,R) \cong \Hom_{R}(R^n,R) \cong \Hom_R(\textstyle\bigoplus_1^nR,R)$$ and that $$R^n \cong \Hom_R(R^n,R) \cong \Hom_R(M \oplus \ker\pi,R).$$
What I want to achive is to see that $\Hom_R(M,R)$ as a right $R$-module is a summand of a free group but for that I need to find the existence of a $R$-morphism not just a $Z(R)$ one, but I don't really know how to proceed, also I haven't used the fact that $M$ is left and $\Hom_R(M,R)$ is right.
Any help will be appreciated.
Since $M$ is finitely generated and projective, we have a split short exact sequence of the form $$0\rightarrow K\rightarrow F\rightarrow M\rightarrow 0$$ where $F$ is a free left $R$-module of finite rank. Since $\operatorname{Hom}_R(\cdot,R)$ is an additive functor, it preserves split short exact sequences. So $$0\rightarrow \operatorname{Hom}_R(M,R)\rightarrow \operatorname{Hom}_R(F,R)\rightarrow \operatorname{Hom}_R(K,R)\rightarrow 0$$ is also a split short exact sequence of right $R$-modules. But $\operatorname{Hom}_R(F,R)$ is a free right $R$-module. This shows $\operatorname{Hom}_R(M,R)$ is projective.