$$\begin{aligned} PF^2 &= AF \cdot PF + PA \cdot PF \\ &= EF^2 + PI \cdot PD. \end{aligned} \qquad \textcircled{1}$$
Since $PQ$ is the diameter of circle $O$ and point $I$ is on $AQ$, we see that $AI \perp AP$. Consequently,
$$\angle IDF = \angle IAP = 90^{\circ}.$$
Thus, we have
$$PF^2 - EF^2 = PD^2 - ED^2.$$
Combining ①, we have
$$PI \cdot PD = PD^2 - ED^2.$$
Thus
$$QD^2 = ED^2 = PD^2 - PI \cdot PD = ID \cdot PD.$$
Consequently, we have
$$\triangle QID \sim \triangle PQD. \qquad ②$$
Since $PQ$ is the diameter of circle $O$, we see that $BP \perp BQ$. Suppose that $PQ$ is the perpendicular bisector of $BC$ at point $M$. Note that $I$ is the incentre of $\triangle ABC$. We have $QI^2 = QB^2 = QM \cdot QP$. Thus,
$$\triangle QMI \sim \triangle QIP. \qquad ③$$
By ② and ③, we see that $\angle IQD = \angle QPD = \angle QPI = \angle QIM$. Hence, $MI \parallel QD$. Let $IK \perp BC$ be at $K$. Then $IK \parallel PM$; thus,
$$\frac{PM}{IK} = \frac{PD}{ID} = \frac{PQ}{MQ}.$$
By the Circle-Power Theorem and the Sine Theorem, we know that
$$\begin{aligned} PQ \cdot IK &= PM \cdot MQ = BM \cdot MC \\ &= \left(\frac{1}{2}BC\right)^2 = (R \sin \angle BAC)^2, \end{aligned}$$
thus, $\sin^2 \angle BAC = \frac{PQ \cdot IK}{R^2} = \frac{2R \cdot r}{R^2} = \frac{2r}{R}$.
$\square$