$H(\kappa)$-absoluteness of a formula Let $\varphi(x,y)$ be an $\in$-formula which is absolute between transitive models of ZF minus powerset axiom. Then $\exists x\, \varphi(x,y)$ is $H(\kappa)$-absolute, where $H(\kappa)$ is the set $\{x\, |\, card(TC(\{x\}))<\kappa\}$. $\kappa$ is uncountable regular. 
This is a homework question and I would prefer a hint instead of a solution.
 A: Assuming ZFC in $V$. Since $\kappa$ is uncountable regular, $H(\kappa)$ models all axioms of ZFC except powerset. (*)
The relativization of $\exists x \, \varphi(x,y)$ to $H(\kappa)$ is $\exists x\in H(\kappa) \,\varphi(x,y)$, $\varphi$ is absolute between $V$ and $H(\kappa)$ because of (*). So it needs to be shown that, if for some $y\in H(\kappa)$ there is $x\in V$ with $\varphi(x,y)$, then there is $x\in H(\kappa)$ with $\varphi(x,y)$.
For $y\in H(\kappa)$, let $T_y:=TC(\{y\})\subseteq H(\kappa)$ with $|T_y|<\kappa$. Let $\kappa'\geq \kappa$ regular be so that if there is $x\in V$ with $\varphi(x,y)$, there is $x\in H(\kappa')$ with $\varphi(x,y)$. Let $S_y$ be a Skolem hull of $T_y$ in $H(\kappa')$ and let $M_y$ be the transitive collapse of $S_y$, with $\pi:S_y\to M_y$ being a $\in\!\!-\!\!\in$ isomorphism. $|S_y|<\kappa$ and $|M_y|<\kappa$. 
Assume there is a $x'\in V$ with $\varphi(x',y)$. Then there is $x\in S_y$ with $\varphi(x,y)$. Since $S_y$ is an elementary substructure of $H(\kappa')$, it is a model of ZF without powerset axiom and $\varphi^{S_y}(x,y)$ holds, because $\varphi$ is absolute. Because $\pi$ is a isomorphism and its restriction to the transitive set $T_y$ is the identity, $\varphi^{M_y}(\pi(x),y)$ holds. Because $M_y$ is isomorphic to $S_y$, ZF without powerset is true in $M_y$ and therefore $\varphi(\pi(x),y)$. 
Now $TC(\{M_y\})=\{M_y\}\cup M_y$ has cardinality less than $\kappa$, so $M_y\in H(\kappa)$. Because $H(\kappa)$ is transitive and $\pi(x)\in M_y$, $\pi(x)\in H(\kappa)$.
