The decreasing order of the ionization potential of the following elements is $N e > C 1 > P > S > Mg > A l .$
In a period, on moving from left to right, ionisation enthalpy increases. In a group, on moving from top to bottom, the ionisation energy decreases.
Closed shell $(Ne)$, half-filled $(P)$ and completely filled configuration $(Mg)$ are the cause of the higher value of I.E.
Since octet is complete in $Ne$, a huge amount of energy is required to remove an electron as it will lead to disruption of stable electronic configuration. $P$ has higher ionisation energy than $S$ as $P$ has a half-filled $3 p$ subshell. Removal of an electron from $P$ atom will break the stability of half-filled subshell.
$Mg$ has higher ionisation energy than $Al$ as $Mg$ has completely filled $3 s$ subshell. Removal of an electron from the $Mg$ atom will break the stability of completely filled subshell. Also, less amount of energy is required to remove an electron from p-subshell than from subshell.
The outer electronic configurations of $Mg$ and $Al$ are $3 s ^{2}$ and $3 s ^{2} 3 p ^{1}$ respectively.