jmc

We can rewrite $Q{P}_b$ as $Q\cdot b+P$, or $\left(\frac{P}{2}\right)b+P$ in base 10. So, we have $P\cdot P=\left(\frac{P}{2}\right)b+P$. The value of $P$ is two less than $b$, so we substitute $(P+2)$ for $b$ and get $$\begin{array}{rl}{P}^2& =\frac{P(P+2)}{2}+P\Rightarrow \\ P^2& =\frac{P^2+2P+2P}{2}\Rightarrow \\ 2P^2& =P^2+4P\Rightarrow \\ P^2-4P& =0\Rightarrow \\ P(P-4)& =0.\end{array}$$That means $P=0$ or $P=4$, but the problem says $P$ is a digit from 1 to 9. So, the value of $P$ is $\boxed{4}$.