9. (15 points) As shown in Figure 3, in $\triangle A B C$, $B C=a, C A=b$, $\angle A C B=60^{\circ}$, $\triangle A B D$ is an equilateral triangle, and $P$ is its center. Find the length of $C P$.
II. 9. Connect $P A$ and $P B$. Then $\angle A P B=120^{\circ}$. Also, $\angle A C B=60^{\circ}$, so points $C$, $B$, $P$, and $A$ are concyclic. Note that, $P A=P B=\frac{2}{3} \times \frac{\sqrt{3}}{2} A B=\frac{\sqrt{3}}{3} A B$.
By Ptolemy's theorem, we have
$$
\begin{array}{l}
P A \cdot C B+P B \cdot A C=A B \cdot C P \\
\Rightarrow \frac{\sqrt{3}}{3} A B(a+b)=A B \cdot C P \\
\Rightarrow C P=\frac{\sqrt{3}}{3}(a+b) .
\end{array}
$$