The CORRECT order of first ionisation enthalpy is:

We recall the general rule that the first ionisation enthalpy, symbolised as $$IE_1$$, increases from left to right across a period because the nuclear charge increases while the principal quantum number $$n$$ remains the same. This means the outermost electrons are held more tightly, so more energy is required to remove one electron.

Moving through the third period we therefore expect $$IE_1(\text{Na}) < IE_1(\text{Mg}) < IE_1(\text{Al}) < IE_1(\text{Si}) < IE_1(\text{P}) < IE_1(\text{S}) \dots$$, but two well-known exceptions disturb this smooth rise. These exceptions arise from extra stability associated with completely filled and half-filled subshells.

First we compare magnesium and aluminium.

We write their electronic configurations in the ground state:

$$\text{Mg : } 1s^22s^22p^63s^2$$

$$\text{Al : } 1s^22s^22p^63s^23p^1$$

Magnesium has a completely filled $$3s$$ subshell ($$3s^2$$). Aluminium’s outer electron is the first electron in the $$3p$$ subshell ($$3p^1$$). A completely filled subshell is especially stable because there is no other orbital of the same energy available to lower the electron-electron repulsion. Removing an electron from magnesium would destroy this stable configuration, so comparatively more energy is required. In contrast, removing the single $$3p$$ electron from aluminium is easier because it restores the stable $$3s^2$$ core. Thus

$$IE_1(\text{Al}) < IE_1(\text{Mg}).$$

Next we compare phosphorus and sulfur. Their configurations are

$$\text{P : } 1s^22s^22p^63s^23p^3$$

$$\text{S : } 1s^22s^22p^63s^23p^4$$

Phosphorus has a half-filled $$3p$$ subshell ($$3p^3$$). A half-filled set of degenerate $$p$$ orbitals is of extra stability because each electron occupies a separate orbital with parallel spins, minimising electron-electron repulsion. Sulfur, on the other hand, already has one of its $$3p$$ orbitals doubly occupied ($$3p^4$$), increasing repulsion. Therefore it is slightly easier to remove an electron from sulfur than from phosphorus. Consequently

$$IE_1(\text{S}) < IE_1(\text{P}).$$

Combining these two anomalies with the usual left-to-right increase gives the order (from lowest to highest) as

$$IE_1(\text{Al}) < IE_1(\text{Mg}) < IE_1(\text{S}) < IE_1(\text{P}).$$

Translating this inequality back into the required arrangement we obtain

Al < Mg < S < P.

This sequence exactly matches Option C in the list provided.

Hence, the correct answer is Option C.