The way you've decided to deform the Kähler form you'll need to check that $\rho$ glues together on coordinate charts, then that the form you've defined is positive-definite (which should be automatic for "small" $\rho$ by compactness). I try to avoid gluing charts when I can, so allow me to address a small deformation of your question.
If you want to deform the Kähler form in such a way that you still get a Kähler form (i.e. the associated metric is Kähler, not merely hermitian$^*$), then there are slightly easier ways to go about this. Denote by $p : \mathbb C^{n+1} \setminus \{0\} \longrightarrow \mathbb P^n$ the canonical projection, then it is known that the pullback of the Fubini-Study metric by $p$ is $p^* \omega = i \partial \bar \partial \log |z|^2$, where $|z^2| = |z_0|^2 + \ldots + |z_n|^2$.
Now, the Hodge number $\dim_{\mathbb C}H^{1,1}(\mathbb P^n, \mathbb C)$ is equal to $1$, so if $\alpha$ is any other Kähler metric on $\mathbb P^n$, then by the $\partial \bar \partial$ lemma there exists a smooth real function $\phi$ on $\mathbb P^n$ such that $\alpha = \omega + i \partial \bar \partial \phi$. Pulling this back via $p$, we have that $p^* \alpha = i \partial \bar \partial (\log |z|^2 + p^*\phi)$. The function $p^* \phi$ comes from $\mathbb P^n$, so it is constant on any line in $\mathbb C^{n+1} \setminus \{0\}$ (i.e. $p^*\phi(\lambda z) = p^*\phi(z)$ for all scalars $\lambda \not= 0$).
You now get any Kähler metric on $\mathbb P^n$ by picking a smooth function $\phi$ on $\mathbb C^{n+1} \setminus \{0\}$ as above and considering the form $p^*\alpha$. As it is constant on any line, then it descends to the projective space. You do need to check that $p^* \alpha$ is positive-definite, which will end up being a condition on $\phi$ (it must be $\omega$-plurisubharmonic), and you should be able to explicit what this condition is by some violent calculations.
$^*$ If you want a non-Kähler metric on $\mathbb P^n$, multiply $\omega$ by any positive non-constant function.