Geometry With Probability

Let $AP = \alpha$ and $PC = 7 - \alpha$ as shown below.

Clearly $\triangle APX \sim \triangle PCY$ . Then

$\begin{aligned} \\ \dfrac{PY}{CY} &= \dfrac{AX}{PX} \\ \\ \dfrac{PY}{(7 - \alpha) \cos 15^\circ} &= \dfrac{\alpha \sin 15^\circ}{PX} \\ \\ PX \cdot PY &= \alpha (7 - \alpha) \sin 15^\circ \cos 15^\circ \qquad \small \color{#3D99F6} \text{[Then, using } \sin \theta \cos \theta = \dfrac{1}{2} \sin 2\theta] \\ \\ &= \dfrac{\alpha}{2} (7 - \alpha) \sin 30^\circ \\ \\ &= \dfrac{\alpha}{4} (7 - \alpha) \\ \\ \end{aligned}$

By the required condition,

$\begin{aligned} \\ PX \cdot PY > 3 \ &\implies \ \dfrac{\alpha}{4} (7 - \alpha) > 3 \\ \\ &\implies \ \alpha (7 - \alpha) > 12 \\ \\ &\implies \ \alpha^2 - 7\alpha + 12 < 0 \\ \\ &\implies \ (\alpha - 4)(\alpha - 3) < 0 \\ \\ &\implies \ 3 < \alpha < 4 \\ \end{aligned}$

Thus $P$ can lie on a segment of length $1$ on $AC$ which has length $7$ , so the desired probability is $\dfrac{a}{b} = \dfrac{1}{7}$ , and our answer is $9ab = \boxed{63}$