Is the first-order theory of complete atomic Boolean algebras finitely axiomatizable? A complete atomic Boolean algebra is one that is isomorphic to a power set algebra, that is, a power set along with the operations of union, intersection, complement, the empty set, and the universal set. Is the first-order theory of the class of complete atomic Boolean algebras finitely axiomatizable? I conjecture that it is, and in fact all you need besides the Boolean algebra axioms is an axiom stating that it is atomic. Is this true? If not, what other axioms do you need?
 A: Yes, the first-order theory of complete atomic Boolean algebras (BAs) is finitely axiomatizable, since it is equal to the theory of atomic BAs, as suggested by bof in the comments. So this theory is axiomatized by the finitely many  axioms for BAs, together with the additional axiom: $$\forall x\, (x = \bot \lor \exists y\, (y \leq x\land \forall z\, (z \leq  y\rightarrow (z = y\lor z = \bot)))).$$
To prove this, we need to show that every atomic BA is elementarily equivalent to a complete atomic BA. This is trivial for finite atomic BAs, since every finite BA is complete. So it remains to show that every infinite atomic BA is elementarily equivalent to a complete atomic BA. This follows from the fact that the theory of infinite atomic BAs is complete, so any infinite atomic BA is elementarily equivalent to any infinite complete atomic BA, such as $\mathcal{P}(\omega)$.
As bof also suggested in the comments, you can prove that the theory of infinite atomic BAs is complete using a back and forth argument / Ehrenfeucht–Fraïssé game. Instead of outlining this argument, I'll just tell you about a much more general theorem: Tarski's complete elementary invariants for Boolean algebras.
Let $B$ be a BA. We say an element $x\in B$ is atomic if for all $y\leq x$ with $y\neq \bot$, there exists an atom $z\leq y$. And we say an element $x\in B$ is atomless if there is no atom $z \leq x$. Let $I(B)$ be the ideal generated by the atomic and atomless elements. That is, $$I(B) = \{y\vee z\mid \text{$y$ is atomic and $z$ is atomless}\}.$$
Now define a sequence of BAs by induction: $B^{(0)} = B$ and $B^{(n+1)} = B^{(n)}/I(B^{(n)})$. Tarski's first invariant $n$ is the minimum natural number such that $B^{(n)}$ is the trivial algebra or $\infty$ if there is no such $n$.
If $n = 0$ (i.e. $B$ is already trivial) or $n = \infty$, then this is the only invariant. Otherwise, $B^{(n)}$ is trivial, but $B^{(n-1)}$ is non-trivial, and we define two more invariants by looking at $B^{(n-1)}$. Tarski's second invariant is just the question of whether $B^{(n-1)}$ is atomic, and Tarski's third invariant is the number of atoms in $B^{(n-1)}$, which could be any natural number or $\infty$ if there are infinitely many.
So for example, any infinite atomic BA has invariants $(1,\text{atomic},\infty)$. A finite BA has invariants $0$ if it is trivial or $(1,\text{atomic},n)$ if it has $n$ atoms. Any atomless BA has invariants $(1,\text{not atomic},0)$.
Now the theorem is that two BAs are elementarily equivalent if and only if they have the same Tarski invariants. The classic reference for this is

Tarski, A., "Arithmetical classes and types of Boolean algebras," Bulletin of the American Mathematical Society, vol. 55 (1949), p. 63.

But I couldn't easily find a copy of this paper online. You can also find a proof in the Handbook of Boolean Algebras, Volume 1, where Section 18 (the first half of Chapter 7) is devoted to the proof.
Since you're interested in the quesiton of finite axiomatization: By looking at the form of the axiomatization of the Tarski invariants, it follows that a completion of the theory of BAs is finitely axiomatizable if and only if none of its invariants are $\infty$.
