(a) Start with the Cartesian equation \(\left( x - \frac{1}{2} \right)^2 + y^2 = \frac{1}{4}\).

Expand and rearrange: \(x^2 - x + \frac{1}{4} + y^2 = \frac{1}{4}\) which simplifies to \(x^2 + y^2 = x\).

In polar coordinates, \(x = r \cos \theta\) and \(x^2 + y^2 = r^2\), so \(r^2 = r \cos \theta\).

Factor to get \(r(r - \cos \theta) = 0\). Since \(r \neq 0\), \(r = \cos \theta\).

(b) Set \(r = \cos \theta = \sin 2\theta\).

Use the identity \(\sin 2\theta = 2 \sin \theta \cos \theta\), so \(\cos \theta = 2 \sin \theta \cos \theta\).

Thus, \(\cos \theta (1 - 2 \sin \theta) = 0\). Since \(\cos \theta \neq 0\), \(\sin \theta = \frac{1}{2}\).

Therefore, \(\theta = \frac{\pi}{6}\) and \(r = \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}\).

The polar coordinates of \(P\) are \(\left( \frac{\sqrt{3}}{2}, \frac{\pi}{6} \right)\).

(c) Sketch the curves \(C_1: r = \cos \theta\) and \(C_2: r = \sin 2\theta\) on the same diagram, marking the point \(P\).

(d) The area \(R\) is given by:

\(\frac{1}{2} \int_0^{\frac{\pi}{6}} \sin^2 2\theta \, d\theta + \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \cos^2 \theta \, d\theta\).

Use identities: \(\sin^2 2\theta = \frac{1}{2}(1 - \cos 4\theta)\) and \(\cos^2 \theta = \frac{1}{2}(1 + \cos 2\theta)\).

Integrate to find:

\(\frac{1}{4} \left[ \theta - \frac{1}{4} \sin 4\theta \right]_0^{\frac{\pi}{6}} + \frac{1}{4} \left[ \theta + \frac{1}{2} \sin 2\theta \right]_{\frac{\pi}{6}}^{\frac{\pi}{2}}\).

Calculate to get \(\frac{1}{4} \left( \frac{\pi}{6} - \frac{\sqrt{3}}{4} \right) + \frac{1}{4} \left( \frac{\pi}{2} - \frac{\pi}{6} - \frac{\sqrt{3}}{4} \right)\).

Simplify to \(\frac{1}{4} \left( \pi - \sqrt{3} \right)\).