In trying to prove the statement in the question (where $P$ is some notion of forcing and $\kappa$ a regular cardinal), I'm stuck in the following step:
Fix a $P$-name $\tau$ and $p\in P$. Put $A_p:=\{\sigma: (\sigma,p)\in\tau\}$. I'd like to produce a set $B_p\subset A_p$ with $|B_p|<\kappa$ and $$\forall \sigma\in A_p \exists \pi\in B_p (p\Vdash \sigma=\pi)$$ As the poset has the $\kappa$-cc, I can find a $\lambda<\kappa$ and a name $\dot f$ with $\Vdash \dot f:\lambda\overset{\text{onto}}{\longrightarrow}\tau$. I also know that $p\Vdash \sigma\in\tau$ whenever $\sigma \in A_p$, so given any $\sigma\in \tau$ I can find an extension $q\le p$ and an ordinal $\alpha<\lambda$ with $q\Vdash \dot f(\alpha)=\sigma$. My problem is that I'd like $p$, rather than some extension, to force such a statement.
If $\tau$ is a set of ordinals, then by $\kappa$-cc, there is some $\alpha<\kappa$ such that $\Vdash\sup\tau < \check\alpha$. Now define $B_\xi=\{p\in P\mid p\Vdash\check\xi\in\tau\}$, then $\{B_\xi\mid\xi<\alpha\}\in H_\kappa$, and so $\overline\tau = \{(\check\xi,p)\mid p\in B_\xi\}\in H_\kappa$.
You can now either use the axiom of choice and the fact that $H_\kappa$ is closed under Mostowski collapses, to conclude the full result. Or you can do this now by recursion on the Kinna–Wagner rank of $\tau$.
That is, sets of ordinals have KW-rank $0$; and $A$ has KW-rank $\alpha$ if $\alpha=\sup\{\text{KW-rank}(a)+1\mid a\in A\}$. Using this idea, define "semi-canonical names" to replace $\check\xi$, and rinse-repeat by induction for $\alpha<\kappa$.