The correct order in which the first ionisation potential increases is:
(a)Na, K, Be
(b)K, Na, Be
(c)K, Be, Na
(d)Be, Na, K

Hint: The amount of energy required to remove an electron from an isolated atom or a molecule is called ionisation energy or ionisation potential. It gives an idea of chemical reactivity of atoms or molecules.

Complete step by step solution:
Ionisation energy or potential is the minimum amount of energy required for an electron to come out of the influence of the nucleus.
It usually increases with increase in atomic number as we move from left to right in a period and decreases for higher energy orbital, i.e down the group.
When we move from top to bottom, that is, down the group, ionisation potential decreases. First ionisation potential is the energy required to remove the first electron or one electron from the valence orbital of the atom or an ion.
Let us consider the electronic configurations of Na, K and Be
\[Na-1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{1}}\]
\[K-1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}4{{s}^{1}}\]
\[Be-1{{s}^{2}}2{{s}^{2}}\]
The first ionisation potential will be more in Be because Be has a stable electronic configuration and is on top of the periodic table, so to remove an electron from its valence orbital requires a lot more energy than Na and K.
The size of K is greater than Na, and also it is present just below Na in its group. Hence, ionisation potential is more for Na when compared with K.

Thus, the correct answer to the question is option (b).

Note: We need to keep in mind that the Beryllium has a stable electronic configuration, it requires more energy to remove an electron from its orbital. Ionisation potential decreases down the group.