9. Let $a$, $b$, $c$ be the roots of the cubic equation
$$
x^{3}+3 x^{2}+5 x+7=0
$$

The cubic polynomial $P$ satisfies
$$
\begin{array}{l}
P(a)=b+c, P(b)=a+c, \\
P(c)=a+b, P(a+b+c)=-16 .
\end{array}
$$

Then $P(0)=$ $\qquad$
9.11.

By Vieta's formulas, we have
$$
a+b+c=-3 \text {. }
$$

Let $Q(x)=P(x)+x+3$, then
$$
\begin{array}{l}
Q(a)=P(a)+a+3 \\
=P(a)-b-c=0, \\
Q(b)=P(b)+b+3 \\
=P(b)-a-c=0, \\
Q(c)=P(c)+c+3 \\
=P(c)-a-b=0 .
\end{array}
$$

Thus, $Q(x)=t(x-a)(x-b)(x-c)$
$$
=t\left(x^{3}+3 x^{2}+5 x+7\right) \text {. }
$$

Notice that, $P(-3)=-16$.
So, $Q(-3)=-8 t=P(-3)-3+3$
$$
=P(-3)=-16 \text {. }
$$

Therefore, $t=2$.
Hence, $P(x)=Q(x)-x-3$
$$
=2\left(x^{3}+3 x^{2}+5 x+7\right)-x-3 \text {. }
$$

Therefore, $P(0)=11$.