 Solution

If $w=1$ in $P$, its image is the identity in the output group, so $c=1$. The equations $u_j^{-1}c{u}_j=g_j$ then give $z=d=1$ and $x_{i}z=1$ for every $i$, hence $x_i=1$. Therefore $C=1$, and the relation $C=a_1$ gives $a_1=1$. Since the normal closure of $a_1$ is $Q$, every $a_i$ is $1$. The remaining identifications make every $u_j$ equal to $1$ as well. All output generators vanish, provingThis proves collapse even when the original group $P$ itself is nontrivial.