And the other directed is easy, the map $f:2^\omega \longrightarrow \mathcal P({}^{<\omega}2)$ with $f(x)=\{\{x|n\}:n\in\omega\}$, then $ A=f^{-1}(\mathcal I_A)$.

I can find a boud rank for $\mathcal I_A$ as follows: Define $X \subseteq \mathcal P({}^{<\omega}2)$ with $x\in X$ iff $\exists n \forall a\in [{}^{<\omega}2]^n((\forall s\in a\Rightarrow s\in x)\Rightarrow (\exists s\not= t\in a (s\prec t \vee t\prec s )))$. Define $g:X\longrightarrow [2^\omega]^{<\omega}$ with $g(x)=\{y\in 2^\omega:\forall n\exists m> n y|m\in x\}$, then $\mathcal I_A=g^{-1}([A]^{<\omega})$, but I can't compute the rank of $\mathcal I_A$ is $\xi$.