Thank you for your help!
I mean that it is zero because we get
$$
g_{ac}g^{ac}g^{bd}=\frac{1}{2}g^{ab}g_{ac}g^{cd} \implies \delta_a^a g^{bd}=\frac{1}{2}g^{b}_{\phantom{b}c}g^{cd}\implies D g^{bd}=\frac{1}{2}g^{bd},
$$
but the dimension is not usually one half :oops:.
I have accidentally derived a very wrong result from the contracted Bianchi identity and I can't see where the error is. I'm sure it's something obvious, but I need someone to point it out to me as I've gone blind. Thanks!
Start with the contracted Bianchi identity,
$$
\nabla_a \left(...