2. As shown in the figure, the diameter $A B=8$ of $\odot O$, and $C$ is a point on $\odot O$ with $\angle C O A=60^{\circ}$. Extend $A B$ to $P$ such that $B P=\frac{1}{2} B O$, and connect $C P$ to intersect the semicircle at point $D$. Draw a perpendicular from point $P$ to $A P$ intersecting the extension of $A D$ at $H$. Then the length of $P H$ is ().
A: $\frac{\sqrt{3}}{3}$
$\mathrm{B}: \frac{\sqrt{3}}{2}$
C: $\frac{2 \sqrt{3}}{3}$
D: $\sqrt{3}$
Answer $\mathrm{C}$.
Analysis Connect $B D, B H$.
Since $A B$ is the diameter, $\angle A D B=90^{\circ}$, so $\angle B D H=90^{\circ}$.
Since $H P \perp A P$ at point $P$, $\angle B P H=90^{\circ}$, that is, $\angle B D H+\angle B P H=180^{\circ}$, so $B, D, H, P$ are concyclic, hence $\angle P B H=\angle P D H$.
Since $\angle C O A=60^{\circ}$, $\angle A D C=30^{\circ}$, thus $\angle P D H=\angle A D C=30^{\circ}$, so $\angle P B H=30^{\circ}$.
Since $A B=8, B P=\frac{1}{2} B O$, $B P=2$, so $P H=2 \cdot \tan 30^{\circ}=\frac{2 \sqrt{3}}{3}$.