The question is whether every extension $\Gamma$ of $G$ by the $G$-module $E$ splits, and whether any two sections of the split extension are conjugate. So, as Geoff says, we have to show that $E$ is complemented in $\Gamma$, and any two complements are conjugate. Here is an elementary argument sent by a correspondent :
Since the quotient $G=\Gamma/E$ is soluble, a minimal normal subgroup of $G$ is an elementary abelian $q$-group $N/E$ for some prime $q$ different from $p$. By Sylow's Theorems there is a complement $Q$ to $E$ in $N$. Consider the normaliser $H$ of $Q$ in $\Gamma$. By the Frattini argument (simply the fact that $Q$ is a Sylow $q$-subgroup of $N$, so $N$ acts transitively, hence $\Gamma$ also acts transitively, by conjugation on the set of conjugates of $Q$ in $N$, which is the set of conjugates of $Q$ in $\Gamma$ since $N$ is normal in $\Gamma$), $\Gamma = H.E$. But the intersection of $H$ and $E$ is trivial (any element of $E$ that normalises $Q$ centralises it since $E$ is normal in $QE$, whereas $E$ is its own centraliser in $\Gamma$). Thus $H$ is a complement of $E$ in $\Gamma$ and moreover, any complement of $H$ in $\Gamma$ has a conjugate of $Q$ as a normal subgroup and therefore is conjugate to $H$. That is to say, there is a unique conjugacy class of complements of $E$ in $\Gamma$ (all of them maximal proper subgroups of $\Gamma$).
Addendum. The correspondent has further provided the following references :
on p.55 of `Finite Soluble Groups' by Doerk & Hawkes (De Gruyter 1992),
on p.102 of `Group Theory II' by Suzuki (Springer 1986), and
on p.159 of `Endliche Gruppen I' by Huppert (Springer 1967).