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 $\Gamma$).