 Solution

Since the two-transitive group action is primitive, the orbits of any nontrivial normal subgroup form only the single block $\Omega$. Thus $N$ is transitive. If $N$ is regular, identify $\Omega$ with $N$ by $x\mapsto x\alpha$. Then $G=N{G}_{\alpha }$, $N\cap G_{\alpha }=1$, and conjugation by $G_{\alpha }$ gives automorphisms of $N$. The two-transitive group action makes these automorphisms transitive on $N\setminus \{1\}$. All nonidentity elements therefore have the same order, which must be a prime $p$: a suitable power of an element of composite order would have a smaller nontrivial order. Cauchy's theorem for finite groups now shows $|N|=p^d$.

The center of a group of prime-power order is nontrivial and is a characteristic subgroup. The transitive conjugation action on nonidentity elements therefore forces $Z(N)=N$. Hence $N$ is an elementary abelian group, which we identify with the additive vector space ${\mathbb{F}}_p^d$. Conjugation by $G_{\alpha }$ is linear on this vector space. It is faithful: an element fixing $\alpha$ and centralizing $N$ fixes every $x\alpha$, hence is the identity permutation. ConsequentlyHere the general affine group acts by translations followed by invertible linear maps.