This is from Kallenberg.
Suppose $(\xi, \eta) \stackrel{d}{=} (\xi, \zeta)$, where $\eta$ is $\zeta$-measurable. Then $\xi$ is conditionally independent of $\zeta$ given $\eta$.
The hint is to first show that $P(\xi \in B | \eta) \stackrel{d}{=} P(\xi \in B | \zeta)$.
I'm guessing I need to use the hypothesis that $\eta = E[ \eta | \zeta ]$ somewhere, but other than that I could use some hints. Thanks!
EDIT: I made some progress. Let $A$ be a set in the underlying sigma field. Then we know that, $$ \begin{align*} E[ E[1_{\xi \in B} | \eta] 1_{ \eta^{-1}(A) } ] &= E[ 1_{\xi \in B} 1_{ \eta^{-1}(A) } ] \\ &= E[ 1_{B}(\xi(\omega)) 1_{A}(\eta(\omega)) ] \\ &= E[1_B(\xi(\omega)) 1_A( \zeta(\omega)) ] \\ &= E[ 1_{\xi \in B} 1_{\zeta^{-1}(A)} ] \\ &= E[ E[1_{\xi \in B} | \zeta] 1_{\zeta^{-1}(A)} ] \end{align*} $$ How do I take this and extend it to show $P(\xi \in B | \eta) \stackrel{d}{=} P(\xi \in B | \zeta)$?