Number of $S_n$-orbits in $P^k(\{1,\dots,n\})$

This is a particular case of a question I asked on Mathematics Stackexchange, question which got no answer so far.

Let $$n$$ and $$k$$ be integers with $$n\ge1$$, $$k\ge0$$, and let $$a(n,k)$$ be the number of orbits of the symmetric group $$S_n$$ on the $$k$$-th iterated power set $$P^k(\{1,\dots,n\})$$ of the set $$\{1,\dots,n\}$$.

We have $$a(n,0)=1$$, $$a(n,1)=n+1$$, $$a(1,k+1)=2^{a(1,k)}$$ and $$\frac{b(n,k)}{n!}\le a(n,k)\le b(n,k),$$ where $$b(n,k)$$ is the cardinality of $$P^k(\{1,\dots,n\})$$.

We also have $$a(2,2)=12$$. Indeed, the twelve orbits of the two-element group $$S_2$$ in $$P^2(\{1,2\})$$ can be described as follows.

The set $$P^2(\{1,2\})$$ having sixteen elements, it suffices to give the eight fixed points. The complement of a fixed point being a fixed point, it suffices to give four fixed points $$F_1,F_2,F_3,F_4$$ such that, for all $$i,j$$, the subset $$F_i$$ is not the complement of $$F_j$$. A possible choice of the $$F_i$$ is $$\varnothing,\ \{\varnothing\},\ \{\{1,2\}\},\ \{\varnothing,\{1,2\}\}.$$

Can anybody compute, say, $$a(2,3),a(3,2)$$ and $$a(3,3)$$?

If you don't feel like computing these three integers, I'd be most grateful if you could nevertheless explain to me a simple (perhaps not efficient) way of computing them.

Edit 1. Here are more details about the equality $$a(2,2)=12$$. We have $$P(\{1,2\})=\{\varnothing,\{1\},\{2\},\{1,2\}\}.$$ So $$P^2(\{1,2\})=P(\{\varnothing,\{1\},\{2\},\{1,2\}\})$$ has $$16$$ elements, one of cardinality $$0$$ (that is, with zero elements), which is the empty set $$\varnothing,$$ four of cardinality $$1$$, which are $$\{\varnothing\},\{\{1\}\},\{\{2\}\},\{\{1,2\}\},$$ six of cardinality $$2$$, which are $$\{\varnothing,\{1\}\}, \{\varnothing,\{2\}\}, \{\varnothing,\{1,2\}\},$$ $$\{\{1\},\{2\}\}, \{\{1\},\{\{1,2\}\}, \{2\},\{\{1,2\}\}$$ four of cardinality $$3$$, which are $$\{\varnothing\}^c,\{\{1\}\}^c,\{\{2\}\}^c,\{\{1,2\}\}^c,$$ where $$A^c$$ denotes the complement of $$A$$ in $$P(\{1,2\})$$, and one of cardinality $$4$$, which is $$P(\{1,2\}).$$ Now let $$X$$ be one of the sixteen above expressions.
If you swap the symbols $$1$$ and $$2$$ in $$X$$, you get either $$X$$ itself, or a different expression $$Y$$, and if you perform the same operation $$Y$$, you get $$X$$ back.
By inspection one sees that there are $$8$$ invariant expressions and $$4$$ pairs of non-invariant expressions, that is $$12$$ "orbits".
Edit 2. An inductive formula for $$a(2,k)$$ is given in an edit to this Mathematics Stackexchange post, but I'm still unable to compute $$a(3,2)$$.