$f:A\rightarrow B$ is a one to one function. Every element in A will have a corresponding element in B. Therefore, the size of range for this is n(A) and n(B) $\geq$ n(A).

$g:2^{A}\rightarrow 2^{B}, g(C)=\left \{f(x) \mid x\in C \right \}$ , since $f$ is one to one, for every subset of A there will be corresponding subset of B. Therefore this is also a one to one function and size of range for this is n($2^{A}$).

$h:2^{B}\rightarrow 2^{A}, h(D)=\left \{ x \mid x\in A, f(x)\in D \right \}$ this function is not a one to one function. Every subset of B will be mapped to subset of A for which it has all the images of subset of A. size of range for this function will be n($2^{A}$)

That said, now $g\left ( h\left ( D \right ) \right )$ will also have the range of size n($2^{A}$). Since $n(A)\leq n(b)$, n($2^{A}$) must be lest than or equal to n($2^{B}$). The answer is $g\left ( h\left ( D \right ) \right )\subseteq D$.

For example let $A = \left\{1, 2 \right\}$ and $B = \left\{a, b, c\right\}$. Let $f(1) = a, f(2) = b$. Now,

$g\left( \left\{ \right\} \right) = \left\{\right\} $

$g\left( \left\{1\right\} \right) = \left\{a\right\} $

$g\left( \left\{2\right\} \right) = \left\{b\right\} $

$g\left( \left\{1,2\right\} \right) = \left\{a,b\right\} $

$h\left( \left\{\right\} \right) = \left\{\right\} $

$h\left( \left\{a\right\} \right) = \left\{1\right\} $

$h\left( \left\{b\right\} \right) = \left\{2\right\} $

$h\left( \left\{c\right\} \right) = \left\{\right\} $

$h\left( \left\{a, b\right\} \right) = \left\{1, 2\right\} $

$h\left( \left\{a, c\right\} \right) = \left\{1\right\} $

$h\left( \left\{b, c\right\} \right) = \left\{2\right\} $

$h\left( \left\{a, b, c\right\} \right) = \left\{1, 2\right\} $

Now we can see that for any $D \subseteq B, g(h(D)) \subseteq D$. Had the function $f$ been bijective (one-one and onto or one-one and co-domain = range) , then we would have got $g(h(D)) = D$.