jmc

Since $P(x)$ is defined only for $x\ge -2$, we must have $G(a)\ge -2$ in order for $P(G(a))$ to be defined. Therefore, we must have $4-3a\ge -2$, so $6\ge 3a$, which means $2\ge a$. Checking, we see that $P(G(2))=P(-2)=4$, so $a=\boxed{2}$ is the largest constant $a$ such that $P(G(a))$ is defined.