You can prove this using your dual version, for instance. By semisimplicity of the ring, we may fix a complementary submodule $N'$ for $N$ within $M$, as well as a complement $P'$ for $P$ within $N$, so that we have $$N = P \oplus P'$$ and $$M = N \oplus N' = P \oplus P' \oplus N'.$$ But then $N'$ is a submodule of $M$ with $N' \cong M/N \cong M$. It follows from the dual version that $P' \oplus N'$ is also isomorphic to $M$, which means that $$M/P \cong P' \oplus N' \cong M.$$

(An alternative approach would be to examine decompositions of the three modules into direct sums of simple submodules. Using direct sum complements as above, one would argue that in nontrivial situations, all simples occur with the same infinite cardinality in the various complemented modules.)