Given that,

$f\left ( p, q \right ) = \underbrace{p\times p\times p\times \cdots\: \cdots\: \cdots \times p}_{q\;\text{terms}} = p^{q}; \;\;f\left ( p, 1 \right )=p$

$g\left ( p, q \right ) = p^{p^{p^{p^{p^{\vdots^\;{\vdots\;^{\vdots^{\text{up to $q$ terms}}}}}}}}};\;\;g\left ( p, 1 \right )=p$

Now, we can check all the options.

$\text{Option A}:\;f(2,2) = g(2,2)$

$\Rightarrow 2 \times 2 = 2^{2^{2}}$

$\Rightarrow {\color{Red}{\boxed{4 = 16\;\text{(False)}}}}$

$\text{Option B}:\;f\left ( g\left ( 2,2 \right ) ,2\right ) < f\left ( 2,g\left ( 2,2 \right ) \right )$

$\Rightarrow f(16,2) < f(2,16)$

$\Rightarrow 16^{2} < 2^{16}$

$\Rightarrow {\color{Green}{\boxed{256 < 65536\;\text{(True)}}}}$

$\text{Option C}:\;g\left ( 2,1 \right ) \neq f\left ( 2,1 \right )$

$\Rightarrow {\color{Red}{\boxed{2 \neq 2 \;\text{(False)}}}}$

$\text{Option D}:\; f\left ( 3,2 \right )> g\left ( 3,2 \right)$

$\Rightarrow 3^{2} > 3^{3^{3}}$

$\Rightarrow {\color{Red}{\boxed{3^{2} > 3^{27}\;\text{(False)}}}}$

Correct Answer $:\text{B}$