B will work out as adequate, by reinterpreting $\land$ in S as an abbreviation for a formula with just the connectives $\lor$ and $\lnot$.
Thus, disjunctive normal forms would work out as having abbreviations in them.
For example, a disjunctive normal form like
((A$\land$B)$\lor$($\lnot$A$\land$B))
would consist of an abbreviation for
($\lnot$($\lnot$A$\lor$$\lnot$B)$\lor$$\lnot$($\lnot$$\lnot$A$\lor$$\lnot$B)).
But, is the above in disjunctive normal form? (my book implies it is not in disjunctive normal form by definition and so does Wikipedia)
If not, then disjunctive normal form is not necessary for proving expressive adequacy.
So, here's another possibility:
Note that every expressive adequacy means we can represent every truth-function. We could represent all truth-functions of an arity n by all possible bits having (2^n) members (by (2^n) I mean 2 to the nth power).
Thus, we have a sort of natural ordering for the outputs of truth-functions as follows:
[0, 1]
[00, 01, 10, 11]
[0000, 0001, 0010, ..., 1110, 1111]
[00000000, 000000001, ..., 11111111]
So, you might try to show that there exists a formula with $\lor$ and $\lnot$ which can always output any of those bits for a truth function of a given arity n, and then have some sort of induction that allows us to move up an arity.
Edit: Post basically shows the key idea in his paper A General Theory of Elementary Propositions.
Roughly, the idea goes that you show that you can obtain the following truth tables:
p F1(p) F2(p) F3(p) F4(p)
0 0 0 1 1
1 0 1 0 1
Then we show that for any truth table where the function has arity n, we can build every truth table for arity (n+1). Post uses the term 'order' for what we call arity.