As shown in Figure 6, let $\angle P C Q = \gamma, \angle P B A = \theta$. From the given conditions, we have
$$\frac{P C}{B P} + 1 = \frac{P C}{P Q}, \text{ i.e., } \frac{P C}{P A} \cdot \frac{P A}{B P} + 1 = \frac{P C}{P Q}.$$

By the Law of Sines, we get
$$\begin{array}{l} \frac{\sin 30^{\circ}}{\sin \gamma} \cdot \frac{\sin \theta}{\sin 30^{\circ}} + 1 \\ = \frac{\sin \left(\theta + 60^{\circ}\right)}{\sin \gamma}, \end{array}$$

which simplifies to
$$\begin{array}{l} \sin \theta + \sin \gamma \\ = \sin \left(\theta + 60^{\circ}\right). \end{array}$$

Thus, $\sin \gamma$
$$\begin{array}{l} = \sin \left(\theta + 60^{\circ}\right) - \sin \theta \\ = 2 \cos \left(\theta + 30^{\circ}\right) \sin 30^{\circ} = \sin \left(60^{\circ} - \theta\right). \end{array}$$

Therefore, $60^{\circ} - \theta > 0^{\circ}$, and $\gamma = 60^{\circ} - \theta$ or $\gamma + (60^{\circ} - \theta) = 180^{\circ}$, i.e.,
$$\gamma + \theta = 60^{\circ} \text{ or } \gamma = \theta + 120^{\circ}.$$

Note that $\angle B P C = \theta + \gamma + 60^{\circ}$. If $\gamma = \theta + 120^{\circ}$, then $\angle B P C = \theta + (\theta + 120^{\circ}) + 60^{\circ} = 2 \theta + 180^{\circ} > 180^{\circ}$, which contradicts $\angle B P C < 180^{\circ}$. Therefore, only $\gamma + \theta = 60^{\circ}$, and in this case, $\angle B P C = 60^{\circ} + 60^{\circ} = 120^{\circ}$.

Let $I$ be the incenter of $\triangle A B C$. Then point $I$ lies on $P A$ or the extension of $A P$, and
$$\begin{array}{l} \angle B I C = 180^{\circ} - \frac{\angle B + \angle C}{2} = 180^{\circ} - \frac{120^{\circ}}{2} \\ = 120^{\circ} = \angle B P C. \end{array}$$

Therefore, point $P$ must coincide with $I$.
Thus, $P$ is the incenter of $\triangle A B C$.