= Solution
If $\alpha=\partial\bar\partial\beta$ and $h\in\mathcal H^{p,q}$, the <Kähler Laplacian identity> makes $h$ both $\partial$- and $\bar\partial$-harmonic. Hence $\partial^*h=\bar\partial^*h=0$, and integration by parts gives $\langle\alpha,h\rangle=0$.
Conversely, let $d\alpha=0$ and $\alpha\perp\mathcal H^{p,q}$. Pure type gives $\partial\alpha=\bar\partial\alpha=0$. <Dolbeault Hodge decomposition> removes the harmonic and coexact components, so
$$
\alpha=\bar\partial\theta,\qquad\theta=\bar\partial^*\gamma.
$$
Now $\bar\partial(\partial\theta)=-\partial\alpha=0$, and the Kähler anticommutation identity gives $\bar\partial^*(\partial\theta)=-\partial(\bar\partial^*\theta)=0$. Thus $\partial\theta$ is $\bar\partial$-harmonic and therefore $\partial$-harmonic; being $\partial$-exact, it vanishes. Moreover $\theta\in\operatorname{im}\bar\partial^*$ is orthogonal to the common $\bar\partial$- and $\partial$-harmonic space. Its $\partial$-Hodge decomposition therefore gives $\theta=\partial\phi$. Hence
$$
\alpha=\bar\partial\partial\phi=-\partial\bar\partial\phi.
$$
Taking $\beta=-\phi$ proves the <harmonic orthogonality criterion for ddbar exactness>
$$
\boxed{\alpha=\partial\bar\partial\beta\Longleftrightarrow\alpha\perp\mathcal H^{p,q}(X).}
$$
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