The problem is to prove that for $u<v\leq a$, then $$P_0 (T_a <t , u <B_t<v) = P_0 (2a-v<B_t<2a-u)$$
The solution first defines $$Y_s(w) = \begin{cases}1& s<t \quad\text{and}\quad u< w(t-s)<v\\0&\text{otherwise}\end{cases}$$. $$ \bar{Y}_s(w)=\begin{cases}1& s<t \quad\text{and}\quad 2a-v< w(t-s)<2a-u\\0&\text{otherwise}\end{cases} $$
What I can't understand is that due to the symmetry of normal distribution, the following holds. $$E_a Y_S = E_a \bar{Y}_S$$ The difference of $w(t-s)$ is same as $v-u$ but how is it related to the symmetry of normal distribution?
Let $\Gamma = (u,v)$. And let $\Theta \subseteq \mathcal C([0,\infty),\mathbb R)\times [0,\infty)$ be such that $(f,\tau)\in\Theta$ if and only if $\tau<t$ and $f(t-\tau)\in\Gamma-a$.
Moreover, let $W_s := B_{s+T_a}-B_{T_a} = B_{s+T_a}-a$. It is a Brownian motion indepedent from $\mathcal F_{T_a}$, and thus independent from $T_a$. This means that $(W,T_a) \stackrel{\mathcal L}= (-W,T_a)$.
Finally, since $2a-\Gamma \subseteq (a,\infty)$, everything I said allows us to obtain the following series of equalities.
$$\mathbb P(T_a<t, B_t \in \Gamma) = \mathbb P(T_a<t, W_{t-T_a} \in \Gamma - a) = \mathbb P((W,T_a)\in \Theta) = \mathbb P((-W,T_a)\in\Theta) = \mathbb P(T_a<t, -W_{t-T_a}\in\Gamma-a) = \mathbb P(T_a<t, B_t \in 2a-\Gamma) = \mathbb P(B_t \in 2a-\Gamma)$$