Superfluous epimorphisms maps maximal submodules into maximal submodules

Question

We define a superfluous epimorphism, $f:P \to M$, if $f$ is a module epimorphism and $\ker(f) \leq P$ is superfluous, that means that if there is $P_{1} \leq P$ such $P_{1} + \ker(f)=P$ then $P_{1}=P$. I want to prove that if there is a maximal submodule $Q \leq P$ then $f(Q)$ is a maximal submodule of $M$. Let say we have $N \leq M$ such $f(Q) \subseteq N \subseteq M$. I want to show that $N=f(Q)$ or $N=M$ so we proceed by supposing both cases. So I got lost around using the preimages to prove this, also don't know how to use the hypothesis of $\ker(f)$ superfluous on $P$. Thanks.