The electronic configuration with the highest ionization enthalpy is -
(A) $\left[ {Ne} \right]3{s^2}3{p^1}$
(B) $\left[ {Ar} \right]3{d^{10}}4{s^2}4{p^3}$
(C) $\left[ {Ne} \right]3{s^2}3{p^3}$
(D) $\left[ {Ne} \right]3{s^2}3{p^2}$

Hint: The electrons in an atom are attracted by the positively charged nucleus. So, to remove an electron from an atom, energy has to be supplied. The energy required to remove an electron from an isolated gaseous atom in its ground state is defined as ionization enthalpy.

Complete answer:
Let us see how it varies across a period or group:
> Variation along a period: The ionization enthalpy increases with increasing atomic number in a period. On moving across a period from left to right, the nuclear charge increases. The atomic size decreases along a period though the main energy level remains the some due to increased nuclear charge & decrease in atomic size, the valence electrons are tightly bound to the nucleus.
Hence, Ionisation enthalpy keeps on increasing.
> Variation down a group: Ionization enthalpy decreases in moving down the group.
In moving from top to bottom, the nuclear charge increase & atomic size also increases due to additional main energy shells (n).
Due to the inner electrons, the outermost electrons are shielded.
Here, the effect of increase in atomic size & the shielding effect is much more prevalent than the increase in nuclear charge. As a result, the electrons are not tightly bound (less firmly held) to the nucleus.
Hence,Ionisation enthalpy gradually decreases.
Among the given options:
$\left[ {Ne} \right]3{s^2}3{p^1} \to Al$
$\left[ {Ar} \right]3{d^{10}}4{s^2}4{p^3} \to As$
$\left[ {Ne} \right]3{s^2}3{p^3} \to P$
$\left[ {Ne} \right]3{s^2}3{p^2} \to Si$
As we move from $Al$ to $Si$ and $P$, ionization enthalpy keeps on increasing due to increasing nuclear charge & decreasing atomic size, ${e^ - }$ is tightly bound.
On moving down, the group i.e. $P \to As$, due to increase in atomic size, the electrons are less firmly held to the nucleus, so removal of an electron is easier. Hence ionisation enthalpy is lower than $P$. So among the given options electronic configuration $\left[ {Ne} \right]3{s^2}3{p^3}$ has the maximum ionisation enthalpy.

Thus the correct answer is (C).

Note: The Ionisation enthalpy increases with increasing atomic number in a period.
Within a group, there is a gradual decrease in ionization enthalpy in moving down the group.