In the given figure, the block of mass m is dropped from the point...

Here, gain in kinetic energy=loss in potential energy

Now loss in potential energy of particle in falling from height $y$ to ${y}_{0}$ is $mg\left(y - {y}_{0}\right)$.

So, $\DeltaKE =$$K - 0 = - \left[mg\left(y - {y}_{0}\right) - mgy\right]$

Thus, kinetic energy at point $B$ is $K = mg{y}_{0}$