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
9.4.

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

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.