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 $\qquad$.
Given that $O M \perp O N$, then $\overrightarrow{M P} \cdot \overrightarrow{P N}=(\overrightarrow{O P}-\overrightarrow{O M})$
$$
\begin{array}{l}
\cdot(\overrightarrow{O N}-\overrightarrow{O P})=-\overrightarrow{O P}^{2}+(\overrightarrow{O M}+\overrightarrow{O N}) \cdot \overrightarrow{O P} \\
=-\overrightarrow{O P}^{2}+2 \overrightarrow{O D} \cdot \overrightarrow{O P}=-(\overrightarrow{O P}-\overrightarrow{O D})^{2}+\overrightarrow{O D}^{2} \leqslant 4,
\end{array}
$$

The equality holds when point $P$ coincides with point $D$.
Therefore, the maximum value of $\overrightarrow{M P} \cdot \overrightarrow{P N}$ is 4.