The equation to show stock levels in week $n$ is:

${\displaystyle U_n=X+(1-\frac{p}{100})U_{n-1}.}$

Substituting for $U_{n-1}$ and simplifying:

${\displaystyle \begin{array}{ccc}\multicolumn{1}{c}{U_n}& =& X+(1-\frac{p}{100})(X+(1-\frac{p}{100})U_{n-2}\\ \multicolumn{1}{c}{}& =& X+(1-\frac{p}{100})X+{(1-\frac{p}{100})}^2{U}_{n-2}.\end{array}}$

Originally there was no stock so the initial condition is $U_0=0$. To make the expressions easier to read we write $s=(1-\frac{p}{100})$.
Writing $U_n$ in terms of $U_0$:

${\displaystyle U_n=X(1+s+s^2+...+s^{n-1})+s^n{U}_0=\sum _{r=0}^{n-1}{\mathrm{Xs}}^r.}$

Summing this geometric series

${\displaystyle U_n=\frac{X(1-s^n)}{1-s}=\frac{100X}{p}(1-{(1-\frac{p}{100})}^n).}$

To find the limit of $U_n$ over a very long period of time, as $(1-p/100)<1$, we have

${\displaystyle \underset{n\to \infty }{lim}{(1-\frac{p}{100})}^n\to 0}$

and so

${\displaystyle \underset{n\to \infty }{lim}{U}_n=\frac{100X}{p}.}$

The company should ensure that they have enough warehouse space for $\frac{100X}{p}$ items. However, the chances are that sales and production will vary during the year, and therefore they should ensure that they have a buffer zone so that they can have more stock if necessary. Also, they are unlikely to be in business for an infinite length of time and therefore should cater said limit to their needs based on the financial situation.