= Solution
For a <differential form of type (p, q)>, $d\alpha=0$ is equivalent to both $\partial\alpha=0$ and $\bar\partial\alpha=0$, because the two resulting types are distinct. Moreover $\partial\bar\partial\beta$ has type $(p,q)$ when $\beta$ has type $(p-1,q-1)$, and
$$
d(\partial\bar\partial\beta)=0
$$
by the square-zero and anticommutation identities. Thus the denominator defining <Bott-Chern cohomology> is a <vector subspace> of its numerator, making the quotient well defined.
<Complex conjugation of differential-form type> sends a $d$-closed $(p,q)$-form to a $d$-closed $(q,p)$-form. For a representative in the denominator,
$$
\overline{\partial\bar\partial\beta}
=\bar\partial\partial\bar\beta=-\partial\bar\partial\bar\beta.
$$
The minus sign does not change the denominator <vector subspace>. Conjugation consequently induces a conjugate-linear bijection $[\alpha]\mapsto[\bar\alpha]$ between the two <Bott-Chern cohomology> spaces; applying it twice is the identity. Equivalently, \b[$\boxed{H^{p,q}_{BC}(X)\cong\overline{H^{q,p}_{BC}(X)}}$] as complex <vector spaces>.
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