Solution:

Let $(a, b, c) := (P(1), P(P(1)), P(P(P(1))))$, so $P(t) = a t^{2} + b t + c$ and we wish to maximize $P(c)$. Then we have that
$$\begin{aligned} a & = P(1) \\ b & = a + b + c \\ c & = P(b) = a b^{2} + b b + c \end{aligned}$$
The first equation implies $c = -b$. The third equation implies $b^{2}(a + 1) = 0$, so $a = -1$ or $b = 0$. If $b = 0$, then $(a, b, c) = (0, 0, 0)$. If $a = -1$, then $b = (-1) b^{2} + (-1) b + (-b)$ or $b = -\frac{1}{3}$, so $c = \frac{1}{3}$ and $(a, b, c) = \left(-1, -\frac{1}{3}, \frac{1}{3}\right)$.

The first tuple gives $P(c) = 0$, while the second tuple gives $P(c) = -\frac{1}{3^{2}} - \frac{1}{3^{2}} + \frac{1}{3} = \frac{1}{9}$, which is the answer.