The inequality can be rewritten as $\left(1-\frac{1}{n}\right)^x\le 1-P.$ We first solve the equation $\left(1-\frac{1}{n}\right)^t= 1-P.$ Take the (natural) logarithm of both sides. We get $t\log\left(1-\frac{1}{n}\right)=\log(1-P).$ Thus $t=\frac{\log(1-P)}{\log\left(1-\frac{1}{n}\right)}.$ Presumably we want $x$ to be an integer. The smallest integer $x$ for which our inequality holds is $\lceil t\rceil$.
It is clear that $x\to \infty$ as $n\to\infty$. However, something interesting happens if we look at $\frac{x}{n}$. We look instead at the closely related $\frac{t}{n}$, which has the same limit. We have $\frac{t}{n}=\frac{\log(1-P)}{n\log\left(1-\frac{1}{n}\right)}.$ Note that the denominator $n\log\left(1-\frac{1}{n}\right)$ is equal to $\log\left(\left(1-\frac{1}{n}\right)^n\right)$. But $(1-1/n)^n\to e^{-1}$ as $n\to\infty$, so the denominator approaches $-1$. It follows that as $n\to\infty$, $\dfrac{t}{n}$ (and therefore $\dfrac{x}{n}$) approaches $-\log(1-P)$.