1. Recall the periodic trend in first ionization enthalpy

   Across Period 3, first ionization energy generally increases from left to right because effective nuclear charge increases.

   Elements given are: $\text{Na},\ \text{Mg},\ \text{Al},\ \text{Si}$

2. Important exception

   Although the general trend is increasing, there is an exception between Mg and Al.

   • Mg: electronic configuration $\text{Mg} = [Ne]3s^2$
   • Al: electronic configuration $\text{Al} = [Ne]3s^2 3p^1$

   The electron removed from Al is a $3p$ electron, which is higher in energy and less tightly held than the $3s$ electron of Mg. Hence: $IE_1(\text{Al}) < IE_1(\text{Mg})$

3. Known approximate first ionization energies

   Standard values are approximately:

   • Na: $496\ \text{kJ/mol}$
   • Mg: $737\ \text{kJ/mol}$
   • Al: $577\ \text{kJ/mol}$
   • Si: $786\ \text{kJ/mol}$

4. Match with options

   So the sequence is: $496,\ 737,\ 577,\ 786$

   This matches Option B.

5. Check all options

   • A: $496, 577, 786, 737$ — incorrect order
   • B: $496, 737, 577, 786$ — correct
   • C: $786, 737, 577, 496$ — completely incorrect
   • D: $496, 577, 737, 786$ — ignores Mg/Al exception

Therefore, the correct answer is B.