II. (25 points) As shown in Figure 1, point $C$ is on the circle $\odot O$ with diameter $AB$. Tangents to $\odot O$ are drawn through points $B$ and $C$, intersecting at point $P$. Connect $AC$. If $OP = \frac{9}{2} AC$, find the value of $\frac{PB}{AC}$.
II. Connect $O C$ and $B C$.
Since $P C$ and $P B$ are tangents to $\odot O$, we have $\angle P O C = \angle P O B$.
Also, $\angle C O B = 2 \angle O A C$, so $\angle P O B = \angle O A C$.
Therefore, $O P \parallel A C$.
From the problem, we know $\angle A C B = \angle O B P = 90^{\circ}$, $\angle P O B = \angle O A C$,
thus $\triangle B A C \sim \triangle P O B \Rightarrow \frac{A C}{O B} = \frac{A B}{O P}$.
$$
\text{Also, } O P = \frac{9}{2} A C, A B = 2 r, O B = r \text{ (where } r \text{ is the radius of } \odot O),
$$

substituting these values, we get
$$
O P = 3 r, A C = \frac{2}{3} r.
$$

In the right triangle $\triangle P O B$, by the Pythagorean theorem, we have
$$
P B = \sqrt{O P^{2} - O B^{2}} = 2 \sqrt{2}.
$$

Therefore, $\frac{P B}{A C} = 3 \sqrt{2}$.