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Michael Albanese
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I think you want an example of a compact complex manifold $X$ and differential forms $\alpha \in \mathcal{E}^{p,q}(X)$ and $\beta \in \mathcal{E}^{p',q'}(X)$ with $(p',q') \neq (p, q)$ such that $[\alpha] = [\beta]$ in de Rham cohomology.


Let $X$ be a compact complex three-dimensional manifold with a non-closed holomorphic one-form $\eta$, i.e. $\eta \in \mathcal{E}^{1,0}(X)$ such that $\overline{\partial}\eta = 0$, but $d\eta = \partial\eta \neq 0$.

Set $\theta = \eta + \bar{\eta}$. Then

$$d\theta = d\eta + d\bar{\eta} = \partial\eta + \overline{\partial}\bar{\eta} = \partial\eta + \overline{\partial\eta} \in \mathcal{E}^{2,0}(X)\oplus\mathcal{E}^{0,2}(X).$$

Note that

$$d(\partial\eta) = \partial(\partial\eta) + \overline{\partial}(\partial\eta) = -\partial\overline{\partial}\eta = 0$$

and $d\overline{\partial\eta} = \overline{d(\partial\eta)} = 0$, so $\alpha := \partial\eta \in \mathcal{E}^{2,0}(X)$ and $\beta := -\overline{\partial\eta} \in \mathcal{E}^{0,2}(X)$ define de Rham cohomology classes $[\alpha], [\beta] \in H^2_{\text{dR}}(X, \mathbb{C})$. Moreover,

$$[\alpha] = [\partial\eta] = [d\theta - \overline{\partial\eta}] = [-\overline{\partial\eta}] = [\beta].$$


In order to successfully use the approach in the previous incarnation of this answer, I wanted to find a $(p,q)$-form $\eta$ which satisfied $\partial\overline{\partial}\eta = 0$ and $d\eta \neq 0$. A non-closed holomorphic form is an example of such a form.

On a compact complex Kähler manifold, every holomorphic form is closed. Without the Kähler hypothesis, one can still show that on a compact complex $n$-dimensional manifold, every holomorphic $(n-1)$-form is closed. Therefore, the first possible example of a non-closed holomorphic form is a holomorphic one-form on a three-dimensional compact complex manifold. Such examples exist as is discussed in Nakamura's Complex Parallelisable Manifolds and Their Small Deformations (I will expand on this later).

Michael Albanese
  • 19.3k
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  • 160