While Gabe's answer is deleted ("One shouldn't try to work in ZF at 5am"), let me work in ZFC.  

(a) The usual middle-thirds Cantor set $C$ is nowhere dense in $\mathbb R$.  It has cardinal $\mathfrak c = 2^{\aleph_0}$.  So every subset of $C$ is again nowhere dense in $\mathbb R$.  Every subset of $C$ has the property of Baire.  There are at least $2^{\mathfrak c}$ sets with the property of Baire.  [In fact, there are exactly $2^{\mathfrak c}$ sets with the property of Baire.]

(b)  Start with any family $\mathscr U$ of sets, and define
$$
\mathcal A(\mathscr U) = \{ \mathcal A(\mathcal X) : 
\mathcal{X} = \langle X_s : s \in {}^{<\omega}\omega\rangle , X_s \in \mathscr U \text{ for all } s \in {}^{<\omega}\omega\} .
$$
The Suslin operation is idempotent.  That is, if $\mathscr U$
is any family of sets, then $\mathcal A(\mathcal A(\mathscr U)) = \mathcal A(\mathscr U)$.  
The family of "Suslin measurable sets" (a.k.a. coanalytic sets) is
$\mathcal A(\mathscr G)$, where $\mathcal G$ is the family of all open subsets of $\mathbb R$.

(c) Let $\mathscr G_0 = \{(a,b) : a,b\in\mathbb Q, a<b\}$.  Then
$\mathcal A(\mathscr G) = \mathcal A(\mathscr G_0)$.  Since $\mathscr G_0$ is countable and ${}^{<\omega}\omega$ is countable, there are at most $\aleph_0 ^{\aleph_0} = \mathfrak c$ Souslin schemes in $\mathscr G_0$.  We conclude that $\mathcal A(\mathscr G_0)$ has cardinal at most $\mathfrak c$.  So there are at most $\mathfrak c$ Suslin measurable sets in $\mathbb R$.  [In fact, there are exactly $\mathfrak c$ Suslin measurable sets.]

(d)  Conclude there is a set with the property of Baire that is not Suslin measurable.