jmc

We can rewrite $31_b$ as $3b+1$ in base 10. So, we have $P\cdot P=3b+1$. The value of $P$ is one less than $b$, so we substitute $(b-1)$ for $P$ and get $$\begin{array}{rl}(b-1{)}^2& =3b+1\Rightarrow \\ b^2-2b+1& =3b+1\Rightarrow \\ b^2-5b& =0\Rightarrow \\ b(b-5)& =0.\end{array}$$That means $b=0$ or $b=5$, but a base of 0 would mean $P=-1$ and the multiplication problem would not be true. So, the base $b$ is $\boxed{5}$.