Is it true that for a noncommutative $R$, a module $P$ is f.g projective iff $\mathsf{hom}(P,R)=P^\ast \cong P$?
Here's what I thought of as a proof: Since $(-)^\ast$ is additive, it preserves biproduct diagrams. The projective modules are the direct summands of free modules, and f.g projective modules can be taken as direct summands of f.g free modules. For finitely generated free modules the result is known. Then write $F=P\oplus Q$ for $F$ f.g free and apply the dualizing functor. Since its additive we have $P\oplus Q=F\cong F^\ast =P^\ast \oplus Q^\ast$ and this means (I think) that $P^\ast \cong P$ as desired. So it seems the only difference from the $P\cong P^{\ast\ast}$ is just that this isomorphism is not natural.
Is the above proof correct? What's the justification for inferring $P\cong P^\ast$ from the biproduct diagrams?
If $R$ is noncommutative and $P$ is a right $R$-module then $\text{Hom}_R(P, R)$ naturally has the structure of a left $R$-module (and vice versa if $P$ is a left $R$-module), so you can't even ask for this isomorphism because the two objects belong to different categories.
But this is still false if $R$ is commutative. Take $R = \mathcal{O}_K$ to be the ring of integers of a number field $K$ whose ideal class group has a nontrivial element of order anything other than two. The corresponding ideal $I$ is f.g. projective and $I^{\ast}$ corresponds to the inverse of $I$ in the ideal class group; moreover, isomorphism of modules corresponds to equality of elements in the ideal class group.
Your proof fails because you don't ask for any compatibility between the isomorphism $F \cong P \oplus Q$ and the isomorphism $F \cong F^{\ast}$.