The standard symplectic form on $\mathbb{C}P^n$ can be constructed as follows. Recall that $\mathbb{C}P^n$ is the quotient of $S^{2n + 1}$ by the multiplication action of $S^1$; in particular, we have a smooth fiber bundle $\pi : S^{2n + 1} \to \mathbb{C}P^n$ with fiber $S^1$. View $S^{2n + 1}$ as a subset of $\mathbb{C}^{n + 1}$, and let $\omega_\mathrm{std}$ be the standard symplectic form on $\mathbb{C}^{n + 1}$. Note that $\omega_\mathrm{std}$ is invariant under the $S^1$-action: indeed, if $\lambda \in S^1$ and $\ell_\lambda$ denotes multiplication by $\lambda$, we have $$ \ell_\lambda^*\omega_\mathrm{std}(X, Y) = \omega_\mathrm{std}((\ell_\lambda)_*X, (\ell_\lambda)_*Y) = \omega_\mathrm{std}(\lambda X, \lambda Y) = \omega_\mathrm{std}(X, Y). $$ Therefore, the restriction of $\omega_\mathrm{std}$ to $S^{2n + 1}$ is invariant under the $S^1$-action, so it descends to a $2$-form $\omega$ on the quotient $\mathbb{C}P^n$. Then $\omega$ is a symplectic form on $\mathbb{C}P^n$.
I am having a bit of trouble justifying that $\omega_\mathrm{std}|_{S^{2n + 1}}$ descends to a form on $\mathbb{C}P^n$. I believe it works if we use the fact that $\omega_\mathrm{std}|_{S^{2n + 1}}$ vanishes on the kernel of $\pi_* : T_pS^{2n + 1} \to T_{\pi(p)}\mathbb{C}P^n$, since we can then define $$ \omega_{\pi(p)}(X, Y) = (\omega_\mathrm{std})_p(X', Y'), $$ where $X', Y' \in T_pS^{2n + 1}$ are any preimages of $X, Y$ under $\pi_*$. I would like to know if it is necessary to use this fact; the source I am reading makes it seem like $\omega_\mathrm{std}|_{S^{2n + 1}}$ being invariant under the action of $S^1$ is enough.