The following is from Rene Schilling's Brownian motion. Let $B_t$ be a one-dimensional Brownian motion and $\tau_b$ the first time it hits $b$. Then we have
$$P(\tau_b \le t, B_t < b) = E(1_{\tau_b \le t} E[1_{(-\infty,0)}(B_t-b)|\mathscr{F}_{\tau_b}^B])=E(1_{\tau_b\le t} E^{B_{\tau_b}} [1_{(-\infty,0)}(B_{t-\tau_b} -b)])= E(1_{\tau_b \le t} E[1_{(0,\infty)} (B_{t-\tau_b})).$$
Note that as it is written, we do not know whether $t-\tau_b$ is positive.In order to make such arguments work we use the following stronger version of the strong Markov property.
Theorem 6.11. Let $B_t$ be a $BM^d$, $\tau$ an $\mathscr{F}_t^B$ stopping time and $\eta\ge \tau$ where $\eta$ is an $\mathscr{F}_{\tau+}^B$ measurable random time. Then we have for all $\omega \in \{\eta < \infty\}$ and all $u \in \mathscr{B}_b(\mathbb{R}^d)$ $$E[u(B_\eta)|\mathscr{F}_{\tau+}^B](\omega)=E^{B_\tau(\omega)}[u(B_{\eta(\omega)-\tau(\omega)}(\cdot))].$$
In the end, we want to show that $P(\tau_b \le t, B_t < b) = P(\tau_b \le t, B_t > b)$. So we need $E(1_{\tau_b \le t} E[1_{(0,\infty)}(B_{t-\tau_b})]) =P(\tau_b\le t, B_t>b)$. But how does this hold?