If a finite group $G$ has an Abelian Sylow $p$-subgroup $P,$ then by transfer, we have $P \cap G^{\prime} \cap Z(G) = 1.$ Your group $N = O_{p}(N) \times H.$ Since $G/N \cong {\rm PSL}(2,p)$ it now follows that $O_{p}(N) \leq Z(G).$ Furthermore, $O_{p}(N)$ is complemented in $G$ by Gaschutz's theorem ($G$ can't have cyclic Sylow $p$-subgroups of order $p^{2},$ again by a transfer argument, since $Z(G)$ contains an element of order $p,$ which lies outside $G^{\prime}).$ The complement $K$ to $O_{p}(N)$ has $H$ as a normal subgroup, wnd indeed $H$ is complemented in $K$ by Schur-Zassenhaus (which I should have explicitly mentioned). A complement $C$ to $H$ in $K$ is a complement to $N$ in $G.$

If a finite group $G$ has an Abelian Sylow $p$-subgroup $P,$ then by transfer, we have $P \cap G^{\prime} \cap Z(G) = 1.$ Your group $N = O_{p}(N) \times H.$ Since $G/N \cong {\rm PSL}(2,p)$ it now follows that $O_{p}(N) \leq Z(G).$ Furthermore, $O_{p}(N)$ is complemented in $G$ by Gaschutz's theorem ($G$ can't have cyclic Sylow $p$-subgroups of order $p^{2},$ again by a transfer argument, since $Z(G)$ contains an element of order $p,$ which lies outside $G^{\prime}).$ The complement $K$ to $O_{p}(N)$ has $H$ as a normal subgroup, and indeed $H$ is complemented in $K$ by Schur-Zassenhaus (which I should have explicitly mentioned). A complement $C$ to $H$ in $K$ is a complement to $N$ in $G.$