Lets take an example..
Suppose A= {1,2,3} . here n=3
Now P(A)= {∅,{1},{2},{3},{1,2},{1,3},{2,3},{1,2,3}}
now C will contain ∅ (empty set) and ,{1,2,3} (set itself) as ∅ is the subset of every set. And every other subset is the subset of {1,2,3}.
now taking subsets of cardinality 1.
We can take any 1 of {1},{2},{3} as none of the set is subset of the other.
Lets take {2}
Now taking the sets of cardinality 2- {1,2},{1,3},{2,3} .
{2}⊂ {1,2} and {2,3} but we can't take both as none of the 2 is subset of the other.
so lets take {2,3}
so C = {∅,{2},{2,3},{1,2,3}}
So if we observe carefully . We can see that we can select only 1 set from the subsets of each cardinality 1 to n .
i.e total n subsets + ∅ = n+1 subsets of A can be there in C
So even though we can have different combinations of subsets in C but maximum cardinality of C will be n+1 only.
So B is the answer.