Solution:

A geometric series with ratio $r$ and sum $1$ has first term $1 - r,$ so its second term is $r(1 - r).$ If the two ratios are $r$ and $s,$ then $r(1 - r) = s(1 - s)$ gives $r - s = r^2 - s^2,$ and since the series are distinct, $r \ne s,$ forcing $s = 1 - r.$

Say the series with ratio $r$ has third term $r^2(1 - r) = \frac{1}{8},$ i.e. $8r^3 - 8r^2 + 1 = 0.$ Substituting $t = 2r$ gives $t^3 - 2t^2 + 1 = (t - 1)(t^2 - t - 1) = 0.$ The root $t = 1$ makes $r = s = \frac{1}{2}$ (the series would coincide), and $r = \frac{1 - \sqrt{5}}{4}$ forces $s = 1 - r \gt 1,$ which diverges. So $r = \frac{1 + \sqrt{5}}{4}.$

The common second term is $$r(1 - r) = \frac{1 + \sqrt{5}}{4} \cdot \frac{3 - \sqrt{5}}{4} = \frac{2\sqrt{5} - 2}{16} = \frac{\sqrt{5} - 1}{8},$$ so $m = 5,$ $n = 1,$ $p = 8,$ and $100m + 10n + p = 518.$