(a) To find the arc length \(AB\), we first need to find the length of \(OA\). Using the cosine rule in triangle \(OAP\):

\(OA^2 = OP^2 + AP^2 - 2 \cdot OP \cdot AP \cdot \cos(\theta)\)

\(OA^2 = x^2 + x^2 - 2x^2 \cos(\theta)\)

\(OA^2 = 2x^2(1 - \cos(\theta))\)

\(OA = x \sqrt{2(1 - \cos(\theta))}\)

Using the identity \(1 - \cos(\theta) = 2\sin^2(\frac{\theta}{2})\), we have:

\(OA = x \sqrt{4\sin^2(\frac{\theta}{2})} = 2x \sin(\frac{\theta}{2})\)

Since \(\sin(\frac{\theta}{2}) = \cos(\frac{\pi}{2} - \frac{\theta}{2})\), we have:

\(OA = 2x \cos(\frac{\pi}{2} - \frac{\theta}{2})\)

Thus, \(OA = 2x \cos \theta\).

The arc length \(AB\) is given by \(\theta \times OA\):

\(AB = \theta \times 2x \cos \theta = 2x\theta \cos \theta\)

(b) The area of the sector \(OAB\) is:

\(\text{Sector area} = \frac{1}{2} \times (2x \cos \theta)^2 \times \theta\)

\(= 2x^2 \cos^2 \theta \times \theta\)

The area of triangle \(OAP\) is:

\(\text{Triangle area} = \frac{1}{2} \times x \times x \times \sin(\pi - 2\theta)\)

\(= \frac{1}{2} x^2 \sin(2\theta)\)

The area of the shaded region \(APB\) is:

\(\text{Area of } APB = \text{Sector area} - \text{Triangle area}\)

\(= 2x^2 \cos^2 \theta \times \theta - \frac{1}{2} x^2 \sin(2\theta)\)

\(= x^2 \left( 2\cos^2 \theta - \frac{1}{2} \sin 2\theta \right)\)