The CORRECT order of first ionisation enthalpy is:

🧠 Two anomalies to apply: $3s^2$ stability beats $3p^1$, and half-filled $3p^3$ beats $3p^4$ Period 3 IE rises overall, but the smooth rise breaks twice. Mg ($3s^2$ filled subshell) > Al ($3p^1$ alone). And P ($3p^3$ half-filled) > S ($3p^4$ paired). Both anomalies must be applied here.

🗺️ Build the order step by step Al has the lone $3p^1$ — easiest to remove. Mg has the filled $3s^2$ — harder than Al, despite Al having one more proton. S has $3p^4$ with one paired orbital — pair repulsion makes it easier than P. P has half-filled $3p^3$ — most stable in this set.

Order (low to high $IE_1$): $\ce{Al} < \ce{Mg} < \ce{S} < \ce{P}$.

Approximate values (kJ/mol): Al = 577, Mg = 737, S = 999, P = 1011.

⚠️ The trap Option (d) places P below S — that follows the "naive" left-to-right rise across Period 3, ignoring the half-filled $p^3$ stability. P > S is one of the two famous Period 3 IE exceptions. If you don't memorize both anomalies (Mg > Al AND P > S), you will trip on this kind of question.

$\boxed{\text{Answer: (c)}}$