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 at least $2^{\mathfrak c}$ 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, 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.

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.