A

Let's analyze the scenario described in the problem.
Step 1: In the given setup, there is a small object P at a height h above the plane mirror located at the bottom of the tank. The observer O is positioned vertically above P, outside the liquid.
Step 2: The image of P formed by the plane mirror will appear to be at the same distance below the mirror as P is above it. Therefore, the height of the image P' below the mirror will also be h.
Step 3: Since the observer O is outside the liquid, the actual distance between P and its image P' needs to be accounted for with the refractive index of the liquid μ.
Step 4: The apparent distance in the liquid will be calculated by considering that light refracts as it enters the liquid from air. The relationship for the apparent depth is given by the formula:
Apparent depth = Real depth / μ.
Step 5: The apparent depth of P in the liquid will be h/μ, and considering the image P' is also at h/μ below the mirror, the apparent distance between P and P' is:
Apparent distance = rac{h}{μ} + rac{h}{μ} = 2 rac{h}{μ}.
Step 6: Finally, from the above calculations, finding the expression leads us to 2 μ h.
Therefore, the answer is A: 2 μ h.