= Solution
For $Y\sim\operatorname{Geom}(p)$, its mean is $1/p$, so the <size-biased probability generating function> is
$$
G_{\widetilde Y}(s)=\frac{p^2s}{(1-(1-p)s)^2}.
$$
Choose \b[$X_1,X_2$ independent with distribution $\operatorname{Geom}(p)$]. Their shifted sum has generating function, initially for $s>0$,
$$
G_{X_1+X_2-1}(s)=\frac1s\left(\frac{ps}{1-(1-p)s}\right)^2=G_{\widetilde Y}(s),
$$
with both sides continuous and zero at $s=0$. Equality of <probability generating functions> proves
$$
\boxed{\widetilde Y\overset d=X_1+X_2-1.}
$$
Equivalently, $X_1+X_2=k+1$ has $k$ possible positive summand pairs, each of probability $p^2(1-p)^{k-1}$, matching the <size-biased geometric distribution> probabilities.
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