${\displaystyle X_{n+1}=X+(1-\frac{p}{100})X_n}$

Put $X_n=Y_n-C$ then

${\displaystyle Y_{n+1}-C=X+(1-\frac{p}{100})(Y_n-C)}$

then choose $C$ such that $-C=X-C(1-\frac{p}{100})$ then $X=\frac{-\mathrm{Cp}}{100}$ and

${\displaystyle Y_{n+1}=(1-\frac{p}{100})Y_n}$

Hence

${\displaystyle Y_{n+1}=(1-\frac{p}{100}{)}^2{Y}_{n-1}=...=(1-\frac{p}{100}{)}^{n+1}{Y}_0}$

and

${\displaystyle X_n=(1-\frac{p}{100}{)}^n{Y}_0+\frac{100X}{p}.}$

Putting $n=0$, $X_0=0$ gives $Y_0=-\frac{100X}{p}$. Hence

${\displaystyle X_n=\frac{100X}{p}(1-(1-\frac{p}{100}{)}^n)\to \frac{100X}{p}}$

as $n\to \infty$ because $(1-\frac{p}{100})<1$.