Fix a tree $ p $ over $ \omega $. Let $ [p] $ denote the set of all branches of $ p $. Given a set of reals $ F \subseteq \omega^\omega $, let $ T(F) := \{ x \mathord{\upharpoonright} n : x \in F \land n < \omega \} $. Clearly, $ T([p]) \subseteq p $ is a maximal pruned subtree of $ p $.
But $ p \mapsto [p] $ is not absolute for transitive models of ZF. So can we construct a maximal pruned subtree in an absolute way?
My idea: Let $ \mathbf{M} $ be a transitive model for ZF. Suppose $ p \in \mathbf{M} $. Now define $ \langle q_n : n < \omega \rangle $ recursively (within $ \mathbf{M} $) by $ q_0 = p $ and $ q_{n + 1} = q_n \setminus \{ s \in q_n : s \text{ is a terminal node of } q_n \} $ for all $ n < \omega $. Let $ p^* := \bigcap \{ q_n : n < \omega \} $. ($ p \mapsto p^* $ should be absolute.)
Does $ \bigl( p^* = T([p]) \bigr)^\mathbf{M} $ hold? And what about $ p^* = T([p]) $ (in $ \mathbf{V} $)?
First of all, one should know about the rank function and the well-founded part of a tree. This is covered in Alexander Kechris's book Classical Descriptive Set Theory in sections I.2.E and I.2.F for example.
Kechris defines $$ \operatorname{WF}_T := \{ s \in T : T_s \text{ is well-founded}\}. $$ for each tree $ T \subseteq A^{<\omega} $.
(Here $ T_s := \{t \in A^{<\omega} : s^\smallfrown t \in T \} $ for each $ s \in T $.)
One easily sees that $ T^* := T \setminus \operatorname{WF}_T $ is the maximal pruned subtree. This is exactly the set of nodes whose ranks are $ \infty := (\max \{ |A|, \aleph_0 \})^+ $.
Remark that "$ R \text{ is a well-founded relation on } A $" is absolute for transitive models of $ \mathrm{ZF} - \mathrm{P} $ ($ \mathrm{ZF} $ without the power set axiom).
So the definition of $ \operatorname{WF}_T $ (and hence of $ T^* $) is also absolute for transitive models of $ \mathrm{ZF} - \mathrm{P} $.