The idea in my previous answer can, I think, be upgraded to solve the whole problem, as follows. Again, fix Skolem functions for $\mathcal A$ as given by the well-ordering $<$, and again let $D$ be a set of fewer than $\kappa$ ordinals $\delta$, each of which is $\sup(\kappa\cap M_\delta)$ for some $M_\delta\prec\mathcal A$ with $|M_\delta|<\kappa$. I need to show that $\sup(D)$ is also in $C_{\mathcal A}$. For each $\delta\in D$, define $N_\delta$ to be the Skolem hull of $\kappa\cap\bigcup_{\xi\in D, \xi\leq\delta}M_\xi$. With appropriate gratitude for the hypothesis that $\kappa$ is regular, note that $N_\delta$ is an elementary submodel of $\mathcal A$ of size $<\kappa$ and that the sequence $\langle N_\delta\rangle$ is an elementary chain. Let $N$ be its union, and note that it, too, is an elementary submodel of $\mathcal A$ of size $<\kappa$. So it suffices to show that $\sup(D)=\sup(\kappa\cap N)$. The $\leq$ direction here is obvious, as $N$ includes $\kappa\cap M_\delta$ for each $\delta\in D$. To complete the proof, suppose the $\geq$ direction failed. Then we would have $\sup(D)<\sup(\kappa\cap N)$, so there would be an ordinal $\alpha\in\kappa\cap N$ with $\sup(D)\leq\alpha$. By construction, we would have some $\delta\in D$ with $\alpha\in N_\delta$, and so $\alpha$ would be of the form $f(\vec\beta)$ for some Skolem function $f$ and some ordinals $\beta_i$ in $\kappa\cap M_{\xi_i}$ for certain $\xi_i\leq\delta$. For each $i$, we have $\beta_i<\xi_i\leq\delta$, and, since there are only finitely many $i$ (as Skolem functions are finitary), we can find $\gamma<\delta$ with all $\beta_i<\gamma$. Increasing $\gamma$ if necessary, we can arrange that $\gamma\in M_\delta$. In $\mathcal A$, we can define the function $g$ sending each ordinal $\nu<\kappa$ to the supremum of all $f(\vec\eta)<\kappa$ for $\eta$ bounded by $\nu$; the values of this function are $<\kappa$ by regularity. As an elementary submodel of $\mathcal A$, $M_\delta$ is closed under $g$ and, in particular, contains $g(\gamma)$. But (again by elementarity) $g(\gamma)$ majorizes $f(\vec\beta)=\alpha>\sup(D)\geq\delta$. That contradicts the fact that $\delta$ is the supremum of $\kappa\cap M_\delta$, and this contradiction completes the proof.