To deform a form need open condition Let $M$ be a connected, compact 2n-manifold and we have a condition as follows: $$\omega^{n-1} \wedge\Omega \wedge \bar{\Omega}\neq0$$ where $\omega$ is Symplectic form and $\Omega$ is closed complex 1-form on $M$. We would change(deform) the $\Omega$ to nearby a form $\Omega’ $ that it is in represent rational cohomology class. This class exist because above condition is open. I can’t understand that why this condition is open? And why is it related to changing $\Omega$?

They mean that the condition that the $2n$-form be nonzero gives an open condition on the vector space of closed complex $1$-forms. This follows immediately because the function $f\colon \mathcal A^1(M)\to\mathcal A^{2n}(M)\cong \mathcal A^0(M)$ given by $f(\Omega) = \omega^{n-1}\wedge\Omega\wedge\bar\Omega$ is continuous. (By $\mathcal A^k(M)$ I mean the space of complex-valued $k$-forms on $M$. Because $M$ is compact, it's easy to put norms on these vector spaces, if you so desire.) Intuitively, this means that if we wiggle $\Omega$ slightly, the wedge product varies only slightly and hence stays nonzero.