Answer: Na < Mg > Al < Si
The electronic configurations are as follows: $_11Na (3s)^1, _12Mg(3s)^2, _13Al (3s)^2 (3p)^1, _14Si(3s)^2 (3p)^2$
The ionization energy of Mg will be larger than that of Na due to fully filled configuration $(3s)^2$
The ionization of Al will be smaller than that of Mg due to one electron extra than the stable configuration but smaller than Si due to increase in effective nuclear charge of Si.
$\Rightarrow$ Na < Mg > Al < Si