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{align*} (b-1)^2&=3b+1\quad\Rightarrow\\ b^2-2b+1&=3b+1\quad\Rightarrow\\ b^2-5b&=0\quad\Rightarrow\\ b(b-5)&=0.\end{align*} 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} $.