Let $E_1$ be the event when randomly selected person said the direction is east for the first time, and $E_2$ be the event when same person said direction is east the second time. We need to find the probability that indeed direction is east given $E_1$ and $E_2$.
$$\begin{align}
P(E \ | \ E_1, E_2) &= \dfrac{P(E, E_1, E_2)}{P(E_1, E_2)} \\
&= \dfrac{P(E, E_1, E_2)}{P(E, E_1, E_2) + P(W, E_1, E_2)} \\
&= \dfrac{P(E, E_1, E_2, T) + P(E, E_1, E_2, K)}{P(E, E_1, E_2) + P(W, E_1, E_2)}\\
&= \dfrac{P(E, E_1, E_2, T) + P(E, E_1, E_2, K)}{P(E, E_1, E_2, T) + P(E, E_1, E_2, K) + P(W, E_1, E_2, T) + P(W, E_1, E_2, K)}
\end{align}$$
Where $E$ is the event that direction is indeed east, similarly $W$ is the event that direction is west. $T$ be the event that the randomly selected person is tourist, similarly $K$ be the event that randomly selected person is kabrastani.
$$\begin{align}
P(E \ | \ E_1, E_2) &= \dfrac{\frac{2}{3}\frac{3}{4}\frac{3}{4} + \frac{1}{3}.0.0}{\frac{2}{3}\frac{3}{4}\frac{3}{4} + \frac{1}{3}.0.0 + \frac{2}{3}\frac{1}{4}\frac{1}{4} + \frac{1}{3}.1.1} = \dfrac{1}{2}\\\end{align}$$
$\textbf{Option (C) is correct}$