6. As shown in the figure, 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. $\frac{2 \sqrt{6}}{3}$
$A B \perp O B \Rightarrow A B \perp$ plane $P O B \Rightarrow$ plane
$P O B \perp$ plane $P A B \Rightarrow O H \perp$ plane $P A B$
$$
\Rightarrow O H \perp P A \Rightarrow H C \perp P A \text {. }
$$

Thus $V_{O-H P C}=\frac{1}{6} P C \cdot C H \cdot O H$
$$
=\frac{1}{3} C H \cdot O H \leqslant \frac{1}{6}\left(C H^{2}+O H^{2}\right)=\frac{2}{3} \text {, }
$$

equality holds when $C H=O H=\sqrt{2}$. Therefore $\angle O P B=30^{\circ} \Rightarrow O B=2 \sqrt{2} \cdot \frac{\sqrt{3}}{3}=\frac{2 \sqrt{6}}{3}$.
So the answer is $D$.