(a) Start with the Cartesian equation \(\left( x - \frac{1}{2} \right)^2 + y^2 = \frac{1}{4}\). Expand and rearrange to get \(x^2 - x + \frac{1}{4} + y^2 = \frac{1}{4}\), which simplifies to \(x^2 + y^2 = x\). In polar coordinates, \(x = r \cos \theta\) and \(x^2 + y^2 = r^2\). Substitute to get \(r^2 = r \cos \theta\), which simplifies to \(r(r - \cos \theta) = 0\). Since \(r \neq 0\), we have \(r = \cos \theta\).

(b) Set \(\sin 2\theta = \cos \theta\). Using the identity \(\sin 2\theta = 2\sin \theta \cos \theta\), we have \(2\sin \theta \cos \theta = \cos \theta\). Assuming \(\cos \theta \neq 0\), divide by \(\cos \theta\) to get \(2\sin \theta = 1\), so \(\sin \theta = \frac{1}{2}\). Therefore, \(\theta = \frac{\pi}{6}\). The polar coordinates of \(P\) are \(\left( \frac{1}{2}, \frac{\sqrt{3}}{6} \pi \right)\).

(c) Sketch the curves \(C_1: r = \cos \theta\) and \(C_2: r = \sin 2\theta\), marking the intersection 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 the identities \(\sin^2 2\theta = \frac{1}{2}(1 - \cos 4\theta)\) and \(\cos^2 \theta = \frac{1}{2}(1 + \cos 2\theta)\) to integrate. The result is \(\frac{1}{4} (\pi - 3\sqrt{3})\).