The correct option is C As<S<P
The increasing order of ionization enthalpy is As<S<P.
On moving from left to right in a period, the ionization energy
increases with few irregularities where half filled and completely
filled orbitals are involved.
On moving down the group, the ionization energy decreases.
Hence, the ionization enthalpy of As is lowest.
The
ionization energy of P is higher than the ionization energy of S as in
P, the electron is to be removed from half filled 3p orbital
whereas in case of S, the electron is to be removed from 3p orbital
containing 4 electrons.