Let $k$ be a field and $N$ a finite group. Let $M$ be a projective indecomposable $kN$-module. Since the algebra $kN$ is symmetric, it follows that the top and bottom composition factors of $M$ are isomorphic. In particular, there is a nonzero endomorphism of
$M$ sending $M$ onto the socle $\operatorname{soc}(M)$.


I cannot see the connection here. How does being symmetric implies composition factors? Any help would be appreciated!



Top and bottom composition factors of $M$ are isomorphic
For every Frobenius algebra $A$ there is a bijection $\pi$ such that $top(P\_i) \cong soc(P\_{\pi (i)})$ when $P\_i$ denote the indecomposable projective $A$-modules.
Being symmetric implies that $A$ is weakly symmetric (meaning that $\pi$ is the identity). Thus top and socle of every $P\_i$ coincide which is what you asked for when I understand your question correct. For proofs and more on this see the book "Frobenius algebras I" in chapter IV. by Skowronski and Yamagata.
When M is an indecomposable projective $A$-module, let $S:=soc(M)$ be the socle of $M$. Then we have a surjective map $M \rightarrow S$ (since top and socle of $M$ coincide) that induces an isomorphism $top(M)=M/rad(M) \rightarrow S$. Thus we have a surjective map $M \rightarrow soc(M)$.


