Let $p$ be an odd prime. Suppose $G$ is solvable and acts as a nonregular and transitive permutation group on $\Omega$ such that each nontrivial element either fixes no point or exactly $p$ points. Also 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 the order of $G$.
Assume we have a maximal normal intransitive subgroup $M$ of index $p$. Let $\Delta$ be the orbit of $M$ containing $\alpha$, and suppose all $p$ fixed point of $G_{\alpha}$ are contained in $\Delta$ and that $M_{\alpha}$ is a $p$-group with $|M_{\alpha}| > p$.
Why is $N_M(M_{\alpha}) = P$ a Sylow $p$-subgroup of $M$?
As $|N_M(M_{\alpha}) : M_{\alpha}|$ equals the number of fixed point of $M_{\alpha}$ on $\Delta$, and we have exactly $p$ such fixed point, we have $|N_M(M_{\alpha}) : M_{\alpha}| = p$, and as $M_{\alpha}$ is a $p$-group, its normalizer in $M$ is also a $p$-group, but why is it a Sylow $p$-subgroup. If not then it would be contained in a Sylow $p$-subgroup $Q$, and $P < N_Q(P) \le Q$. But I do not see where this would be a contradiction?