A potential barrier of 250 meV exists across a p - n junction. A...

In a $p$-$n$ junction, the $n$-side is at a higher electric potential than the $p$-side due to the built-in potential barrier. The electric field in the depletion region is directed from the $n$-side to the $p$-side.

A hole acts as a positive charge. When it approaches the junction from the $n$-side and moves towards the $p$-side, it moves in the direction of the electric field. Thus, it moves from a higher potential to a lower potential.

The loss in potential energy of the hole is equal to the potential barrier energy, which is $250 \text{ meV}$. By the conservation of energy, this decrease in potential energy results in an equal increase in its kinetic energy.

Initial kinetic energy of the hole, $K_i = 300 \text{ meV}$
Energy gained by the hole $= 250 \text{ meV}$

Final kinetic energy of the hole, $K_f = K_i + \text{Energy gained}$
$K_f = 300 \text{ meV} + 250 \text{ meV} = 550 \text{ meV}$