2. As shown in Figure 1, given that $P A$ and $P B$ are tangent to $\odot O$ at points $A$ and $B$, $P C$ satisfies $A B \cdot P B - A C \cdot P C = A B \cdot P C - A C \cdot P B$, and $A P \perp P C$, $\angle P A B = 2 \angle B P C$. Then $\angle A C B =$ $\qquad$
2. $30^{\circ}$.

As shown in Figure 3.
$$
\begin{array}{l} 
\text { From } A B \cdot P B- \\
A C \cdot P C \\
= A D \cdot P C- \\
A C \cdot P B,
\end{array}
$$

we get
$$
\begin{array}{l}
(A B+A C) \cdot \\
(P B-P C)=0 .
\end{array}
$$

Thus, $P B=P C$.
Also, $P A=P B$, so
$$
P A=P B=P C \text {. }
$$

Therefore, points $A$, $B$, and $C$ lie on a circle with $P$ as the center and $P A$ as the radius.
Let $\angle A C B=\alpha$. Then
$$
\begin{array}{l}
\angle A P B=2 \alpha, \angle B P C=90^{\circ}-2 \alpha . \\
\text { Hence } \angle P A B=2 \angle B P C=180^{\circ}-4 \alpha . \\
\text { Also } \angle P A B=\angle P B A=\frac{1}{2}\left(180^{\circ}-\angle A P B\right) \\
=90^{\circ}-\alpha,
\end{array}
$$

Thus, $180^{\circ}-4 \alpha=90^{\circ}-\alpha$.
Therefore, $\alpha=30^{\circ}=\angle A C B$.