9. Given that $M N$ is a moving chord of the circumcircle of equilateral $\triangle A B C$ with side length $2 \sqrt{6}$, $M N=4$, and $P$ is a moving point on the sides of $\triangle A B C$. Then the maximum value of $\overrightarrow{M P} \cdot \overrightarrow{P N}$ is $\qquad$.
9.4 .

Let the circumcenter of $\triangle A B C$ be $O$.
It is easy to find that the radius of $\odot O$ is $r=2 \sqrt{2}$.
Also, $M N=4$, so $\triangle O M N$ is an isosceles right triangle, and
$$
\begin{array}{l}
\overrightarrow{O M} \cdot \overrightarrow{O N}=0,|\overrightarrow{O M}+\overrightarrow{O N}|=4 . \\
\text { Let }|\overrightarrow{O P}|=x . \\
\text { Then } \sqrt{2} \leqslant x \leqslant 2 \sqrt{2}, \text { and } \\
\overrightarrow{M P} \cdot \overrightarrow{P N}=(\overrightarrow{O P}-\overrightarrow{O M}) \cdot(\overrightarrow{O N}-\overrightarrow{O P}) \\
=-\overrightarrow{O P}+(\overrightarrow{O M}+\overrightarrow{O N}) \cdot \overrightarrow{O P}-\overrightarrow{O M} \cdot \overrightarrow{O N} \\
=-x^{2}+(\overrightarrow{O M}+\overrightarrow{O N}) \cdot \overrightarrow{O P} \\
\leqslant-x^{2}+|\overrightarrow{O M}+\overrightarrow{O N}||\overrightarrow{O P}| \\
=-x^{2}+4 x=4-(x-2)^{2} \leqslant 4,
\end{array}
$$

The equality holds if and only if $x=2$ and $\overrightarrow{O P}$ is in the same direction as $\overrightarrow{O M}+\overrightarrow{O N}$.
Therefore, the maximum value of $\overrightarrow{M P} \cdot \overrightarrow{P N}$ is 4.