= Solution
Taking $x=y=a$ in (c) and using that $B_t$ has no atom at $a$ gives
$$
\mathbb P(T_a\leq t)=\mathbb P(S_t\geq a)=2\mathbb P(B_t\geq a)
=2\left(1-\Phi\left(\frac a{\sqrt t}\right)\right),\qquad t>0.
$$
This is the <distribution function> of the <Brownian first-passage time>; (b) ensures that it is a proper <probability distribution>. Comparing these <distribution functions> gives the <Brownian scaling> identity $T_{ca}\overset d=c^2T_a$ for $c>0$.
For an integer $n\geq1$, set $T_0=0$ and successively hit the levels $a,2a,\ldots,na$. The <Strong Markov property> at these finite <stopping times> and spatial translation imply that
$$
T_a-T_0,\ T_{2a}-T_a,\ldots,T_{na}-T_{(n-1)a}
$$
are <independent random variables>, each with the <probability distribution> of $T_a$. Thus, for <independence>[independent] copies $T_a^{(j)}$,
$$
T_a^{(1)}+\cdots+T_a^{(n)}\overset d=T_{na}\overset d=n^2T_a.
$$
Therefore
$$
\boxed{\frac{T_a^{(1)}+\cdots+T_a^{(n)}}{n^2}\overset d=T_a,\qquad \alpha=\frac12.}
$$
The paper's definition is the <strictly stable distribution> convention, with no centering term, a special case of a <stable distribution>.
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