Recall the trend of first ionization enthalpy

First ionization enthalpy generally:

• increases across a period from left to right,
• decreases down a group.

There are some exceptions due to electronic configuration, shielding, and poor shielding by $d$-electrons.

Check option A

Option A: The first ionization enthalpy of K is less than that of Na and Li.

Alkali metals are: $\text{Li},\ \text{Na},\ \text{K},\ \text{Rb},\ \text{Cs}$

Down the group, ionization enthalpy decreases: $\text{Li} > \text{Na} > \text{K}$

Hence, $IE_1(\text{K}) < IE_1(\text{Na}) \text{ and } IE_1(\text{Li})$

So A is correct.

Check option B

Option B: Xe does not have the lowest first ionization enthalpy in its group.

Group 18 elements are: $\text{He},\ \text{Ne},\ \text{Ar},\ \text{Kr},\ \text{Xe},\ \text{Rn}$

Ionization enthalpy decreases down the group, so among these, Rn has lower first ionization enthalpy than Xe.

Therefore Xe indeed does not have the lowest value in its group.

So B is correct.

Check option C

Option C: The first ionization enthalpy of element with atomic number 37 is lower than that of the element with atomic number 38.

Atomic numbers:

• $37 = \text{Rb}$
• $38 = \text{Sr}$

In period 5, ionization enthalpy generally increases from left to right. Since Rb is an alkali metal and Sr is an alkaline earth metal: $IE_1(\text{Rb}) < IE_1(\text{Sr})$

So C is correct.

Check option D

Option D: The first ionization enthalpy of Ga is higher than that of the d-block element with atomic number 30.

Here:

• $\text{Ga}$ has atomic number $31$
• d-block element with atomic number $30$ is $\text{Zn}$

Compare configurations: $\text{Zn}: [Ar]3d^{10}4s^2$ $\text{Ga}: [Ar]3d^{10}4s^24p^1$

The electron removed from Ga is a $4p$ electron, which is easier to remove than the $4s$ electron of Zn. Hence, $IE_1(\text{Ga}) < IE_1(\text{Zn})$

Therefore the statement "The first ionization enthalpy of Ga is higher than that of Zn" is incorrect.

So D is incorrect.

Final conclusion

The incorrect statement is: $\boxed{\text{D}}$