(a) To find \(\angle ABC\), we can use trigonometry. Since \(P\) is the foot of the perpendicular from \(C\) to \(AB\), \(AP = 12\) cm and \(BP = 3\) cm. Therefore, \(\tan \angle ABC = \frac{9}{3} = 3\). Using the inverse tangent function, \(\angle ABC = \tan^{-1}(3) \approx 1.249\) radians, which rounds to \(1.25\) radians to 3 significant figures.

(b) To find the area of the shaded region, we first find \(BC\) using the sine rule: \(BC = \frac{9}{\sin 1.25} \approx 9.4869\) cm.

The area of sector \(BCQ\) is \(\frac{1}{2} \times (BC)^2 \times \angle ABC = \frac{1}{2} \times 9.4869^2 \times 1.25 \approx 56.207\) square cm.

The area of triangle \(PBC\) is \(\frac{1}{2} \times 9 \times 3 = 13.5\) square cm.

Therefore, the area of the shaded region is \(56.207 - 13.5 = 42.7\) square cm (or \(42.8\) square cm depending on rounding).