In the range $n \le jp^{1/2} < n+1$, the number of $j$'s contributing to the sum is $|A|p^{-1/2} +o(p^{-1/2})$. So that portion of the sum satisfies
$$
\sum_{j=np^{-1/2}}^{(n+1)p^{-1/2}} p(1-p)^j \chi_I(j)= (|A|p^{1/2}+o(p^{1/2})) (1-p)^{np^{-1/2}} ;
$$
here, I can treat $(1-p)^j$ as a constant when doing the sum, at the expense of a multiplicative error $(1-p)^{p^{1/2}}$, which can be absorbed into the other error term, from the ergodic theorem.

So the whole sum equals
$$
|A|p^{1/2}(1+o(1)) \sum_{n\ge 0} (1-p)^{np^{-1/2}}   = (1+o(1))|A| \frac{p^{1/2}}{1-(1-p)^{p^{-1/2}}} \to |A| ,
$$
as desired.