Answer: 406

Solution:

Completing the square, the condition reads $$\begin{aligned} (x-1)^2 + 1 &\le P(x) \\ &\le 2(x-1)^2 + 1. \end{aligned}$$ At $x = 1$ both bounds equal $1,$ so $P(1) = 1.$ The quadratic $P(x) - \left((x-1)^2 + 1\right)$ is nonnegative for all $x$ and vanishes at $x = 1,$ so $x = 1$ is a double root: $P(x) = a(x-1)^2 + 1$ for some constant $a.$

From $P(11) = 100a + 1 = 181$ we get $a = \frac{9}{5}.$ Then $$\begin{aligned} P(16) &= \frac{9}{5} \cdot 225 + 1 \\ &= 405 + 1 = 406. \end{aligned}$$