The excess pressure inside an air bubble of radius r just below the surface of water is P1. The excess pressure inside a drop of the same radius just outside the surface is P2. If T is surface tension then

To determine the relationship between the excess pressures, let us analyze the physical conditions inside an air bubble in water and a water droplet in air.

First, let us consider an air bubble of radius _r just below the surface of water.
An air bubble inside a liquid has only one liquid-gas interface (the inner surface of the bubble).
The excess pressure $P_1$ inside a spherical bubble with a single interface is given by the formula:
$P_1=\frac{2T}{r}$

Next, let us consider a water droplet of the same radius _r in air.
Similarly, a liquid droplet has only one interface (the outer surface of the droplet).
The excess pressure $P_2$ inside a droplet with a single interface is given by the formula:
$P_2=\frac{2T}{r}$

Comparing the two expressions, we see that both the air bubble inside water and the water droplet in air have the same excess pressure since both have only one free surface:
$P_1=P_2$

Therefore, the correct relation is $P_1=P_2$.