Given below are two statements :
Statement I : The second ionization enthalpy of B, Al and Ga is in the order of $$B > Al > Ga$$.
Statement II : The correct order in terms of first ionization enthalpy is $$Si < Ge < Pb < Sn$$.
In the light of the above statements, choose the correct answer from the options given below:

First recall the definition: the $$n^{\text{th}}$$ ionization enthalpy is the energy required to remove the $$n^{\text{th}}$$ electron after the previous $$(n-1)$$ electrons have already been removed. Hence for second ionization enthalpy we must look at the species $$M^{+}$$, not the neutral atom.

Statement I: “The second ionization enthalpy of $$B,\;Al$$ and $$Ga$$ is in the order $$B \gt Al \gt Ga$$”.
Electronic configurations and their singly-charged cations are

$$B : 1s^{2}\;2s^{2}\;2p^{1} \;\;\;\; \rightarrow \;\;\; B^{+} : 1s^{2}\;2s^{2}$$
$$Al : [Ne]\;3s^{2}\;3p^{1} \;\;\;\; \rightarrow \;\;\; Al^{+} : [Ne]\;3s^{2}$$
$$Ga : [Ar]\;3d^{10}\;4s^{2}\;4p^{1} \;\;\;\; \rightarrow \;\;\; Ga^{+} : [Ar]\;3d^{10}\;4s^{2}$$

In $$B^{+}$$ the next electron has to be removed from the compact, inner $$2s$$ orbital which experiences a very high effective nuclear charge, giving a very large $$IE_2$$.
In $$Al^{+}$$ the electron is removed from $$3s$$, farther from the nucleus and better shielded, so $$IE_2$$ is lower.
In $$Ga^{+}$$ the electron is taken from $$4s$$, but the presence of the filled $$3d^{10}$$ core gives poor shielding, so its $$IE_2$$ is higher than that of $$Al^{+}$$ but still lower than that of $$B^{+}$$.

Actual experimental values (kJ mol-1): $$IE_2(B)=2427,\; IE_2(Al)=1817,\; IE_2(Ga)=1979$$.
Order: $$B \gt Ga \gt Al$$, not $$B \gt Al \gt Ga$$. Hence Statement I is false.

Statement II: “The correct order of first ionization enthalpy is $$Si \lt Ge \lt Pb \lt Sn$$”.
Move down group 14 (Si, Ge, Sn, Pb): the atomic size increases, so ionization enthalpy generally decreases; however, poor shielding by $$d$$ and $$f$$ electrons causes a small rise from Sn to Pb.

Experimental $$IE_1$$ values (kJ mol-1):
$$Si=786,\; Ge=762,\; Sn=709,\; Pb=715$$

So the increasing (lowest → highest) order is
$$Sn \lt Pb \lt Ge \lt Si$$, while the decreasing (highest → lowest) order is
$$Si \gt Ge \gt Pb \gt Sn$$.

The sequence given in Statement II, $$Si \lt Ge \lt Pb \lt Sn$$, is exactly opposite to the facts, therefore Statement II is also false.

Because both statements are incorrect, the correct choice is:
Option B which is: Both Statement I and Statement II are false.