Let $\rho$ be a representation of a finite group $G$, and let $P$ and $Q$ be probabilities on $G$. Prove $\widehat{P*Q}(\rho)=\hat{P}(\rho)\hat{Q}(\rho)$
I know that by definition, $\hat{P}(\rho)=\sum_{s}P(s)\rho(s)$, $\hat{Q}(\rho)=\sum_{t}Q(t)\rho(t)$ and $P*Q(\rho)=\sum_{t\in G}P(st^{-1})Q(t)=\sum_{s\in G}P(t)Q(t^{-1}s)$, but I'm not sure how to manipulate this to let both side equal.
Notice
$$ \widehat{P\ast Q}(\rho)=\sum_{s} (P\ast Q)(s)\rho(s)=\sum_s\left(\sum_{ab=s} P(a)Q(b)\right)\rho(s)=\sum_{a,b} P(a)Q(b)\rho(ab). $$
Then, using $\rho(ab)=\rho(a)\rho(b)$, we can factore this double summation as a product of sums:
$$ =\left(\sum_a P(a)\rho(a)\right)\left(\sum_b Q(b)\rho(b)\right)=\widehat{P}(\rho)\widehat{Q}(\rho). $$