$Hom_R(P,P)$ is projective if $P$ is projective.

For the moment I only know a proof for $P$ being finitely generated and projective. Im still posting this, since this answer might help anyways. You can use the result that $text{Hom}_R(bigoplus_{i in I} M_i,N) cong prod_{i in I}text{Hom}_R(M_i,N).$ Since $P$ is projective there exists a set $I$ and a $R$-Module $N$ such that $$P oplus N cong bigoplus_{i in I}R.$$

This gives you $$text{Hom}_R(P,P) oplus text{Hom}_R(N,P) cong text{Hom}_R(P oplus N,P) cong text{Hom}_R(R^{(I)},P) cong prod_{i in I} text{Hom}_R(R,P) cong prod_{i in I} P. $$ Now, as far as I know, direct products of projective modules are not necessarily projective, however in the case where $P$ is finitely generated, we would have a finite direct sum, e.g. $$text{Hom}_R(P,P) oplus text{Hom}_R(N,P) cong bigoplus_{i=1}^n P,$$ thus $text{Hom}_R(P,P)$ is projective too. I hope that this might help, despite it only proving a special case.

Edit: The direct sum is finite in this case, since there exists a $n in mathbb{N}$ and a surjective $R$-linear map, $varphi$, such that the sequence $$ 0 xrightarrow{} text{ker}(varphi) xrightarrow{} R^n xrightarrow{} P xrightarrow{} 0$$ is exact. Since $P$ is projective this sequence splits and we obtain $P oplus text{ker}(varphi) cong R^n$. If we now set $N=text{ker}(varphi)$ we obtain a finite direct sum.