A water drop is divided into $8$ equal droplets. The pressure difference between the inner and outer side of the big drop will be

Volume is conserved when the water droplet is divided into 8 smaller droplets.

Let R and r be the radii of the bigger droplet and smaller droplets respectively.

 From volume conservation:

$\frac{4}{3}{R}^3=8\times \frac{4}{3}{r}^3$

$\Rightarrow r=\frac{R}{2}$

The pressure difference between surfaces is $\Delta P=\frac{4T}{r}$ where T is the surface tension and r is the radius of the drop( since there are two surfaces for a water, it is multiplied by an extra factor of 2 )

Pressure difference from droplet of radius R $P_R=\frac{4T}{R}$

Pressure difference from droplet of radius r $P_r=\frac{4T}{r}=\frac{4T}{\frac{R}{2}}=2\times \frac{4T}{R}=2P_R$

$\Rightarrow P_R=\frac{1}{2}{P}_r$