For this problem, we just take an example which satisfies all the conditions which are given in the Qs.
For example let A={1,2} and B={ a,b,c}. Let f(1) =a, f(2)=b. Now,
g({})={}
g({1})={a}
g({2})={b}
g({1,2})={a,b}
h({})={}
h({a})={1}
h({b})={2}
h({c})={}
h({a,b})={1,2}
h({a,c})={1}
h({b,c})={2}
h({a,b,c})={1,2}
we can see that h is not One-to-One.
Now, we find out g(h(D)).
D =2^{B} = {{},{a},{b},{c} ,{a,b} ,{a,c},{b,c},{a,b,c} }
h(D)= { {},{1},{2},{1,2} }
g({})={}
g({1})={a}
g({2})={b}
g({1,2})={a,b}
g(h(D)) = {{},{a},{b},{a,b}}
Now we can see that for any D⊆B,g(h(D)) ⊆ 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.
The correct answer is,(A) g(h(D))⊆D