Griffith and Harris state on page $116$ that for a closed form $\mu$ on a Kahler manifold of type $(p,q)$ we have $$\mu=\mathcal{H}(\mu)+dd^*G(\mu)$$ Here $$\mathcal{H}:\Omega^{p,q}(M)\to\mathcal{H}^{p,q}(M)$$ is the projection onto the harmonic part, using the Hodge theorem. $G:\Omega^{p,q}(M)\to\Omega^{p,q}(M)$ is the Green's operator.
However, here they seem to use that first, since the metric is Kahler $$\mu=\mathcal{H}(\mu)+\Delta_{\bar\partial}G(\mu)=\mathcal{H}(\mu)+\Delta_{d}G(\mu)$$ Then they must expand: $$\Delta_dG(\mu)=(dd^*+d^*d)G(\mu)=dd^*G(\mu)+d^*dG(\mu)$$ And now they seems to use $$dG=Gd$$ to get rid of the term $d^*dG(\mu)$. However, the Hodge theorem only gives us $$\bar\partial G=G\bar\partial, \bar\partial^* G=G\bar\partial^*$$ So I don't see how we can conclude that $dG=Gd$?