jmc

Rewriting the given quadratics in vertex form, we have $$1+(x-1{)}^2\le P(x)\le 1+2(x-1{)}^2.$$Both of those quadratics have vertex at $(1,1)$; considering the shape of the graph of a quadratic, we see that $P$ must also have its vertex at $(1,1)$. Therefore, $$P(x)=1+k(x-1{)}^2$$for some constant $k$. Setting $x=11$, we have $181=1+100k$, so $k=\frac{9}{5}$. Then $$P(16)=1+\frac{9}{5}\cdot 15^2=\boxed{406}.$$