Example 2 (2004 National High School Competition Question) As shown in Figure $7-3$, the axis section of a cone with vertex $P$ is an isosceles right triangle, $A$ is a point on the circumference of the base circle, $B$ is a point inside the base circle, $O$ is the center of the base circle, $A B \perp O B$, with the foot of the perpendicular at $B, O H \perp P B$, with the foot of the perpendicular at $H$, and $P A=4, C$ is the midpoint of $P A$. When the volume of the tetrahedron $O-H P C$ is maximized, the length of $O B$ is ( ).
A. $\frac{\sqrt{5}}{3}$
B. $\frac{2 \sqrt{5}}{3}$
C. $\frac{\sqrt{6}}{3}$
D. $\frac{2 \sqrt{6}}{3}$
Solution: Choose D. Reason: Since $A B \perp O B, A B \perp O P$, then $A B \perp P B$. Also, $O H \perp P B$, so plane $P A B \perp$ plane $P O B$, thus, $O H \perp H C, O H \perp P A$. Since $C$ is the midpoint of $P A$, then $O C \perp P A$, so, $P C$ is the height of the tetrahedron $P-H O C$, and $P C=2$.

In the right triangle $\triangle O H C$, $O C=2$, so when $H O=H C$, the area $S_{\triangle H X}$ is maximized, which also means the volume $V_{O-H P C}=V_{P H C O}$ is maximized. At this time, $H O=\sqrt{2}$, so $H O=\frac{1}{2} O P$. Therefore, $\angle H P O=30^{\circ}$, so, $O B=O P \cdot \tan 30^{\circ}=\frac{2 \sqrt{6}}{3}$.