Let $P$ be a projective module and $\operatorname{rad}(P)$ be its radical. $\operatorname{rad(P)}$ is a proper submodule. That is, we claim that $P$ has a maximal submodule.
My attempt: We know that for an arbitrary right $R$-module $M$, $MJ \subseteq \operatorname{rad}(M)$, where $J=\operatorname{rad}(R_R)$. Also, if $M$ is a projective module, then $MJ = \operatorname{rad}(M)$. Fix a nonzero element $p \in P$ and consider this $R$-linear map $\phi_{p}:R \longrightarrow P$ such that $\phi_{p}(r)=pr$. Then $\phi_p(J)=PJ=\operatorname{rad}(P)$. Now assume that $\operatorname{rad}(P)=P$, namely, P has no maximal submodule. Then, $\phi_p(J)=P$. Does it give a contradiction ?