6 The vertex of a circular cone is $P$, and its axial section 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 $P A$. Then, 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) $\frac{2 \sqrt{6}}{3}$
Since $A B \perp O B, A B \perp O P$, it follows that $A B \perp P B$, and plane $P A B \perp$ plane $P O B$. Also, since $O H \perp P B$, it follows that $O H \perp H C, O H \perp P A$.
Furthermore, because $C$ is the midpoint of $P A$, it follows that $O C \perp P A$, thus $P C$ 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 $S_{\triangle H O C}$ is maximized, and thus the volume $V_{O-H P C}=V_{P-H C O}$ is maximized. At this point, $H O=\sqrt{2}$, hence $H O=\frac{1}{2} O P$, so $\angle H P O=30^{\circ}, O B=$ $O P \cdot \tan 30^{\circ}=\frac{2 \sqrt{6}}{3}$.
Therefore, the answer is D.