Let $G$ be a transitive permutation group on $\Omega$ which fulfills the following property (P)
(P) each nontrivial element fixes no point or exactly $p$ points.
for some odd prime $p$. Further suppose that for $g \notin N_G(G_{\alpha})$ we have $$ G_{\alpha} \cap G_{\alpha}^g = 1 $$ for some point stabilizer $G_{\alpha}$ and that $p$ is the smallest prime dividing the order of $G$. Assume $G$ to be solvable and that $M$ is a maximal intransitive normal subgroup. We have $|G / M| = q$ where $q$ is prime.
So $M$ has $q$ orbits and $M_{\alpha} = G_{\alpha}$. Let $\Delta$ be the orbit of $M$ containing $\alpha$ and suppose $\Delta$ contains all $p$ fixed points of $G_{\alpha}$. Then $M$ in its action on $\Delta$ and the other orbits has property (P). Further $M$ can have at most $p+1$ representations as a group which has the property (P) with mutually disjoint point stabilizers.
Why does the property that $p$ is the smallest prime dividing $|G|$ implies that $q = p$? And why can $M$ have at most $p+1$ representations as a group with property (P)?
I cannot connect the number $p+1$ of representations to the conlusion that $q = p$. Hopeing someone could clarify this!?