In which of the following pairs, the ionisation energy of the first species is less than that of the second

From the above option we can see that the Atomic Number of Sulphur(S) is 16.

Electronic configuration of sulphur is $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^4$

We can remove the electron from the 3p orbital of the sulphur atom very easily.

It shows first ionisation energy of sulphur is very less.

Now, We know that Atomic number of phosphorus(P) is 15 and Electronic configuration of phosphorus is $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^3.$

It is difficult to remove electron from the 3p orbital of the phosphorus atom.Because it has half-filled electronic configuration.

Therefore, the first  ionisation energy of sulphur is less than that of the phosphorus.

Hence, option(d) is correct.