= Solution
For the <near-origin probability in a maximal-angular-momentum hydrogen state>, integrate the normalized radial probability density:
$$
P(r<r_p)=|A|^2\int_0^{r_p}r^{2l+2}e^{-2r/[a(l+1)]}\,dr.
$$
Throughout this small interval the exponential is $1+O(r_p/a)$ for fixed $l$. Hence
$$
P(r<r_p)=\frac{|A|^2r_p^{2l+3}}{2l+3}
\left[1+O(r_p/a)\right].
$$
Using the normalization above,
$$
\boxed{P(r<r_p)=
\frac{2^{2l+3}}{(2l+3)!(l+1)^{2l+3}}
\left(\frac{r_p}{a}\right)^{2l+3}
\left[1+O(r_p/a)\right],\qquad
C=(2l+3)!(l+1)^{2l+3}.}
$$
The high power of the small radius reflects the near-origin behavior $R\propto r^l$ of the <hydrogen atom> state.
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