1. Locate the elements in Period 3

The elements are all from the 3rd period:

$\text{Mg},\ \text{Al},\ \text{Si},\ \text{P},\ \text{S}$

Their electronic configurations are:

• $\text{Mg}: [Ne]3s^2$
• $\text{Al}: [Ne]3s^2 3p^1$
• $\text{Si}: [Ne]3s^2 3p^2$
• $\text{P}: [Ne]3s^2 3p^3$
• $\text{S}: [Ne]3s^2 3p^4$

2. General trend of first ionization enthalpy across a period

Across a period, first ionization enthalpy generally increases from left to right because effective nuclear charge increases and atomic size decreases.

So the rough order should be:

$\text{Mg} < \text{Al} < \text{Si} < \text{P} < \text{S}$

But we must check the known exceptions.

3. Important exceptions

(i) $\text{Al}$ vs $\text{Mg}$

• In $\text{Mg}$, the electron removed is from $3s$.
• In $\text{Al}$, the electron removed is from $3p$.

Since a $3p$ electron is higher in energy and less tightly held than a $3s$ electron, it is easier to remove from Al.

Therefore:

$\text{Al} < \text{Mg}$

(ii) $\text{S}$ vs $\text{P}$

• $\text{P}: 3p^3$ is a half-filled configuration, which is extra stable.
• $\text{S}: 3p^4$ has one paired electron in a $p$ orbital, so electron-electron repulsion makes removal easier.

Therefore:

$\text{S} < \text{P}$

4. Combine the trend and exceptions

Using the general increase across the period, with the above exceptions:

$\text{Al} < \text{Mg} < \text{Si} < \text{S} < \text{P}$

5. Check options

• A: $\text{Al} < \text{Mg} < \text{Si} < \text{S} < \text{P}$ ✓
• B: $\text{Al} < \text{Mg} < \text{S} < \text{Si} < \text{P}$ ✗ because $\text{Si} < \text{S}$
• C: $\text{Mg} < \text{Al} < \text{Si} < \text{S} < \text{P}$ ✗ because $\text{Al} < \text{Mg}$
• D: $\text{Mg} < \text{Al} < \text{Si} < \text{P} < \text{S}$ ✗ for both exceptions

Hence, the correct option is:

$\boxed{\text{A}}$