$\begin{array}{llll}\text{ Al }& \text{ Si }& \text{ P }& \text{ S }\\ 2,8,3& 2,8,4& 2,8,5& 2,8,6\end{array}$
since ionization energy increases across the period, one would expect the correct order of increasing ionization energy to be:
$Al<Si<P<S$
However, this is not correct!!! The ionization energy of phosphorus is greater than that of Sulphur. This is because phosphorus has a more stable electronic structure
${}_{15}P=1s^{2}2s^{2}2p^{6}3s^{2}3p^3$
${}_{16}S=1s^{2}2s^{2}2p^{6}3s^{2}3p^4$
Note that fully filled and half filled orbitals have higher stability, compare to partially filled orbitals. As P can accommodate a total number of 6 electrons, $p^6$ and $p^3$ will be more stable than $p^1,p^2,p^4\mathrm{\&}{p}^5$. Thus, the correct trend is: $Al<Si<S<P$