The statement that any extension of $G$ by $V$ splits is proved in a joint paper of mine with Tim Dokchitser, the proof starts in the last paragraph of page 12. Our $G$ is slightly different from yours, but the argument carries over completely unchanged. The basic idea is to compute the group $H^2(G,V)$ by hand, which classifies extensions of $G$ by $V$ up to splitting. (In our notation $V$ is $W$ and $G$ is $Q=C\rtimes P$).

Edit: The argument is so short that I might as well give it here.

First, note that $V^H$ is trivial. Indeed, since $H$ is normal, $V^H$ is a subrepresentation. But $C$ is a $p$-group, so it has a fixed vector in $V^H$, which is then fixed under all of $G$. This contradicts irreducibility of $V$ (unless $V$ is the trivial representation, in which case the statement is obvious).

Next, we sandwich $H^2(G,V)$ using the inflation-restriction exact sequence. The following is exact: $$ H^2(G/H,V^H)\longrightarrow H^2(G,V)\longrightarrow H^2(H,V). $$ Now, the leftmost term is 0, since $V^H$ is, and the rightmost term is 0, since it is killed by $|H|$ and $|V|$, which are coprime. So $H^2(G,V)=0$, as required.

The following is a direct proof that any extension of $G$ by $V$ splits. It is taken from a joint paper of mine with Tim Dokchitser, where the proof starts in the last paragraph of page 12.

First, note that for any $n\in\mathbb{N}$, $H^n(H,V)=0$, since it is killed by $|H|$ and $|V|$, which are coprime. This implies that the inflation map $$ H^2(G/H,V^H)\longrightarrow H^2(G,V) $$ is an isomorphism. But also, either $V$ is trivial, in which case what we want to prove is obvious, or $V^H=0$. Indeed, since $H$ is normal, $V^H$ is a subrepresentation. But $C$ is a $p$-group, so it has a fixed vector in $V^H$, which is then fixed under all of $G$. This contradicts irreducibility of $V$, unless $V$ is the trivial representation. So $H^2(G,V)=0$, as required.

The statement that any extension of $G$ by $V$ splits is proved in a joint paper of mine with Tim Dokchitser, the proof starts in the last paragraph of page 12. Our $G$ is slightly different from yours, but the argument carries over completely unchanged. The basic idea is to compute the group $H^2(G,V)$ by hand, which classifies extensions of $G$ by $V$ up to splitting. (In our notation $V$ is $W$ and $G$ is $Q=C\rtimes P$).

The statement that any extension of $G$ by $V$ splits is proved in a joint paper of mine with Tim Dokchitser, the proof starts in the last paragraph of page 12. Our $G$ is slightly different from yours, but the argument carries over completely unchanged. The basic idea is to compute the group $H^2(G,V)$ by hand, which classifies extensions of $G$ by $V$ up to splitting. (In our notation $V$ is $W$ and $G$ is $Q=C\rtimes P$).

Edit: The argument is so short that I might as well give it here.

First, note that $V^H$ is trivial. Indeed, since $H$ is normal, $V^H$ is a subrepresentation. But $C$ is a $p$-group, so it has a fixed vector in $V^H$, which is then fixed under all of $G$. This contradicts irreducibility of $V$ (unless $V$ is the trivial representation, in which case the statement is obvious).

Next, we sandwich $H^2(G,V)$ using the inflation-restriction exact sequence. The following is exact: $$ H^2(G/H,V^H)\longrightarrow H^2(G,V)\longrightarrow H^2(H,V). $$ Now, the leftmost term is 0, since $V^H$ is, and the rightmost term is 0, since it is killed by $|H|$ and $|V|$, which are coprime. So $H^2(G,V)=0$, as required.