$\mathbf{Geometrical \ approach}:$ Each point on the surface of a sphere is at an equal distance (equal to radius) from the center this results in the minimum surface area for a given volume. This can be proved analytically by comparing the surface area of a sphere with that of any other geometrical shape for a given volume.
For example, let's compare the surface areas of a sphere & a cube for a given volume say $V$ of the drop,
For the sphere of a radius $r$ $$\frac{4\pi}{3}r^3=V\implies r=\left(\frac{3V}{4\pi}\right)^{1/3}$$
surface area of the sphere $\color{red}{S_1}=4\pi r^2=4\pi\left(\frac{3V}{4\pi}\right)^{2/3}=\color{red}{\left(\frac{9}{2\sqrt \pi}\right)^{1/3}V^{2/3}\approx1.36 V^{2/3}}$
For a cube with edge length $a$
$$a^3=V\implies a=V^{1/3}$$
surface area of the cube $\color{blue}{S_2}=6 a^2=\color{blue}{6V^{2/3}}$
Comparing surface areas, $\color{red}{S_1}<\color{blue}{S_2}$ i.e. the surface area of a sphere is smaller than that of a cube for a given volume
Similarly, we can analytically compare surface area of a sphere with that of any other geometrical shape.
The minimum surface area of the sphere results in the minimum surface energy of the drop.
that's why the drop takes spherical shape to minimize its potential energy