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.

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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 $d\theta \neq 0$ ($d\theta = 0$ if and only if $\partial\eta = 0$). Furthermore, $\partial\eta$ is $d$-closed as 

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

and as $d\overline{\partial\eta} = \overline{d(\partial\eta)} = 0$, $\overline{\partial\eta}$ is also $d$-closed. Therefore $\alpha := \partial\eta$ and $\beta := -\overline{\partial\eta}$ define de Rham cohomology classes $[\alpha], [\beta] \in H^2_{\text{dR}}(X, \mathbb{C})$. Note however that

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