This is from excercise 7.4.3.
For $u<v\leq a$ , the following holds. \begin{equation} P_0 (T_a<t,u<B_t<v) = P_0 (2a-v<B_t<2a-u) \end{equation}
Letting $(u,v)$ shrink down to $x$, for $a<x$ $$P_0 (T_a<t, B_t = x) = p_t (0,2a-x)$$
I know this should hold intuitively, but is there more rigourous proof of this?
I guess it is due to continuity of Brownian motion, but can't think of formal proof.
It's due to the continuity of $y\mapsto p_t(0,y)$: $$ \lim_{u\uparrow x,v\downarrow x}{P_0(2a-v<B_t<2a-u)\over v-u}=\lim_{u\uparrow x,v\downarrow x}\int_{2a-v}^{2a-u}{p_t(0,y) dy\over v-u}=p_t(0,2a-x). $$