Across period 3, first ionisation enthalpy generally rises as we move from sodium to argon because the nuclear charge keeps increasing while electrons are added to the same shell. But there is a well known dip after every half filled or fully filled subshell is crossed, and that dip is the key to this question.
The four configurations given belong to silicon $(3p^2)$, phosphorus $(3p^3)$, sulfur $(3p^4)$ and aluminium $(3p^1)$. Phosphorus has its three 3p electrons singly occupying all three p orbitals, a symmetric, half filled arrangement. Such an arrangement resists losing an electron because both exchange energy and orbital symmetry are maximised, so phosphorus needs unusually high energy to be ionised.
Sulfur, right next to phosphorus, actually has a slightly lower first ionisation enthalpy than phosphorus even though it carries one more proton, because removing an electron from sulfur relieves the repulsion between the two electrons paired up in one 3p orbital. The same reasoning rules out silicon and aluminium, whose ionisation enthalpies are lower still.
So among the four, the half filled $3p^3$ configuration of phosphorus needs the most energy to remove an electron. The correct choice is option (2).