Let's solve the problem step by step:

1. Find the Eigenvalues of $P$:

The transition matrix $P$ is:

$P = \begin{pmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{4} & \frac{3}{4} \end{pmatrix}$

The eigenvalues $\lambda$ are obtained by solving $\det(P - \lambda I) = 0$:

$\det \left( \begin{pmatrix} \frac{1}{2} - \lambda & \frac{1}{2} \\ \frac{1}{4} & \frac{3}{4} - \lambda \end{pmatrix} \right) = 0$

$\det \left( \begin{pmatrix} \frac{1}{2} - \lambda & \frac{1}{2} \\ \frac{1}{4} & \frac{3}{4} - \lambda \end{pmatrix} \right) = \left( \frac{1}{2} - \lambda \right) \left( \frac{3}{4} - \lambda \right) - \frac{1}{8} = 0$

Simplify:

$\left( \frac{1}{2} - \lambda \right) \left( \frac{3}{4} - \lambda \right) = \frac{3}{8} - \frac{5}{4}\lambda + \lambda^2$

$\lambda^2 - \frac{5}{4}\lambda + \frac{1}{2} = 0$

Solve using the quadratic formula:

$\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \quad a = 1, \, b = -\frac{5}{4}, \, c = \frac{1}{2}$

$\lambda = \frac{-(-\frac{5}{4}) \pm \sqrt{\left(-\frac{5}{4}\right)^2 - 4(1)\left(\frac{1}{2}\right)}}{2(1)}$

$\lambda = \frac{\frac{5}{4} \pm \sqrt{\frac{25}{16} - 2}}{2}$

$\lambda = \frac{\frac{5}{4} \pm \sqrt{\frac{25}{16} - \frac{32}{16}}}{2}$

$\lambda = \frac{\frac{5}{4} \pm \sqrt{-\frac{7}{16}}}{2}$

So, the eigenvalues are ( \lambda