\begin{equation*}
\sum_{x=1}^{\infty} {f(x)} = \sum_{x=1}^{\infty} {(1-p)^{x-1} p} \\ = p \sum_{j=0}^{\infty} {(1-p)^j} = p \frac{1}{1-(1-p)} = 1
\end{equation*}
using a change of variables j = x-1 and the known value for the sum of the geometric series.