 Solution

Killing equation is ${\nabla }_{(a}{K}_{b)}=0$. For the stress-energy current from a Killing vector $J^a=T^a{}_b{K}^b$,The first term vanishes on shell, while the second contracts the symmetric tensor $T^{ab}$ with the antisymmetric derivative selected by Killing's equation. Thus $\boxed{{\nabla }_a{J}^a=0}$.