We can rewrite $ QP_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{align*} P^2&=\frac{P(P+2)}{2}+P\quad\Rightarrow \\ P^2&=\frac{P^2+2P+2P}{2}\quad\Rightarrow \\ 2P^2&=P^2+4P\quad\Rightarrow \\ P^2-4P&=0\quad\Rightarrow \\ P(P-4)&=0.\end{align*} 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} $.