The correct answer is (a) 5.
The question is asking for the value of $n$ in the following equation:
$$P_n = \frac{1}{3}P_{n-1}$$
We can solve this equation by starting with $P_0 = 1$ and then repeatedly substituting the previous value of $P_n$ into the equation. This gives us the following sequence:
$$P_0 = 1$$
$$P_1 = \frac{1}{3}P_0 = \frac{1}{3}$$
$$P_2 = \frac{1}{3}P_1 = \frac{1}{9}$$
$$P_3 = \frac{1}{3}P_2 = \frac{1}{27}$$
$$P_4 = \frac{1}{3}P_3 = \frac{1}{81}$$
As you can see, the sequence is decreasing by a factor of $\frac{1}{3}$ each time. This means that the value of $n$ must be a multiple of 3, since $P_n$ is always a multiple of $\frac{1}{3}$. The only multiple of 3 in the answer choices is 5, so the correct answer is (a) 5.
The other answer choices are incorrect because they are not multiples of 3. For example, 7 is not a multiple of 3, so $P_7$ cannot be equal to $\frac{1}{3}P_{7-1}$.