Exercise IV.6.11 of Kunen says there exists an atomless ccc Baire poset iff there is a Suslin tree; I think here Baire can be taken to mean either countable intersection of dense open sets is dense or not addng $\omega$-sequence, since the poset is atomless. Below is my proof. For simplicitly let's work with a complete Boolean algebra $B$ that is atomless, ccc, and $\omega$-distributive, namely
$$\bigwedge_{n\in\omega}\bigvee_{i\in I}b_{n,i}=\bigvee_{f:\omega\rightarrow I}\bigwedge b_{n,f(n)}$$
We build a Suslin tree $T$ inside $B$ by recursion; the $\alpha$-th level $\mathcal{L}_\alpha(T)$ will be a maximal antichain in $B$. Let the maximal element $1_B$ be the root. Suppose we have built $\mathcal{L}_\alpha(T)$. Take a nontrivial partition (a maximal antichain below $b$ that is not $\{b\}$) of each $b\in\mathcal{L}_\alpha(T)$ and let $\mathcal{L}_{\alpha+1}(T)$ be their union. At limit stage take infimum along all branches of the tree constructed thus far; for some branch the infimum could be zero, but because of $\omega$-distributivity, the collection of all these infima is a maximal antichain. Continue this for $\omega_1$ steps. I think I can show that the result $T$ is a Suslin tree, and moreover $T$ is a complete sub-poset of $B^+$, the set of nonzero elements of $B$ viewed as a poset.
Question: is $T$ a dense subset of $B^+$?
Edit: As pointed out in the comment, $T$ may not be dense; there remains the question of whether one can find a $T$ that is dense, i.e., whether every ccc Baire forcing is equivalent to a Suslin tree. The answer is (sort of trivially) consistently yes, because it may happen that neither exist. If $B$ has a dense subset of size $\aleph_1$, then it's easy to find a dense $T$; so for example, at least assuming CH, the iteration of two Suslin trees is equivalent to a Suslin tree. Now this is starting to look like a duplicate of the MO thread linked in the comment...