9. As shown in Figure 1, in $\triangle A B C$, $\angle A B C=60^{\circ}$, point $P$ is a point inside $\triangle A B C$ such that $\angle A P B=\angle B P C=\angle C P A$, and $P A=8, P C=6$. Then $P B=$ $\qquad$
$9.4 \sqrt{3}$.
It is known that $\angle A P B=\angle B P C=\angle C P A=120^{\circ}$.
$$
\begin{array}{l}
\because \angle P A B=180^{\circ}-120^{\circ}-\angle P B A \\
=60^{\circ}-\angle P B A=\angle P B C \text {. } \\
\therefore \triangle P A B \backsim \triangle P B C, \frac{P A}{P B}=\frac{P B}{P C} \text {. } \\
\end{array}
$$

Therefore, $P B=\sqrt{P A \cdot P C}=\sqrt{6 \times 8}=4 \sqrt{3}$.