Let $G$ be a solvable, nonregular and transitive permutation group acting on $\Omega$ such that each nontrivial element either fixes no point or exactly $p$ points for some prime $p$. And suppose that for $g \notin N_G(G_{\alpha})$ we have $$ G_{\alpha} \cap G_{\alpha}^g = 1. $$ Suppose that $p$ is the smallest prime dividing $|G|$. Let $M$ be a maximal normal subgroup of $G$. Suppose $|G / M| = p$ and that $M$ is intransitive, then $M$ has $p$ orbits and by maximality $M_{\alpha} = G_{\alpha}$. Let $\Delta$ be the orbit of $M$ containing $\alpha$, and suppose that $\Delta$ contains all $p$ fixed points of $G_{\alpha}$. Further assume $L$ is a normal subgroup of $M$ which is regular on $\Delta$ and that $|M_{\alpha}| = p$ and $O_{p'}(G) = 1$.
Why then $|M / L| = p$ and it follows that $G$ has a normal subgroup $S$ such that $|G / S| = p^2$ and $S \le M$?
I see that the set of fixed points of $G_{\alpha}$ is a minimal block, and by the assumption that $\Delta$ contains all of them we must have $|\Delta| = k\cdot p$. Also as $L$ acts regular on $\Delta$, we have $|L| = |\Delta|$, and $M_{\alpha}$ operates equivalent on $L$ by conjugation as on $\Delta$. Also if $|M / L| = p$, then $|G : L| = |G : M||M : L| = p^2$, but $L$ need not be normal in $G$, so the subgroup $S$ need to be constructed in some other way?