 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 $(p-1,q-1)$, andby 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,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, $\boxed{H_{BC}^{p,q}(X)\cong \overline{H_{BC}^{q,p}(X)}}$ as complex vector spaces.