Consider a standard Wiener process $W_t$ with $$ W_0=0,\quad\mathbb E(W_t)=0,\quad \mathbb E(W_t\cdot W_s)=\min(t,s) $$ and let $k\geq0$ be the threshold of interest.
Define the first passage time as $$ \tau_k = \inf\{t\ :\ W_t = k\}. $$ Due to continuity of $W_t$, $$ \mathbb P(W_t > k) = \mathbb P(\{W_t > k\}\cap\{ \tau_k < t\}. $$$$ \mathbb P(W_t > k) = \mathbb P(\{W_t > k\}\cap\{ \tau_k < t\}). $$ Now note that $$ \mathbb P(\{W_t > k\}\cap\{ \tau_k < t\} =\mathbb P(W_t >k\ |\ \tau_k < t)\cdot\mathbb P(\tau_k < t) = \frac12\mathbb P(\tau_k < t) $$$$ \mathbb P(\{W_t > k\}\cap\{ \tau_k < t\} =\mathbb P(W_t >k\ \mid\ \tau_k < t)\cdot\mathbb P(\tau_k < t) = \frac12\mathbb P(\tau_k < t) $$ due to the symmetrical nature of the Wiener process (intuitively, if the process is already at $k$ at some point of time, it can end up below or above $k$ later on with equal probabilities).
Hence, the probability that threshold $k$ will be crossed before time $t$ is equal to $$ \mathbb P(\tau_k < t) = 2\cdot\mathbb P(W_t > k) = 2\cdot\Phi\\left(-\frac{k}{\sqrt{t}}\\right),\quad k>0, $$$$ \mathbb P(\tau_k < t) = 2\cdot\mathbb P(W_t > k) = 2\cdot\Phi\left(-\frac{k}{\sqrt{t}}\right),\quad k>0, $$ where $\Phi$ is a c.d.f. of the standard normal distribution. Using the similar argument for $k<0$, we can conclude $$ \mathbb P(\tau_k < t) = 2\cdot\Phi\\left(-\frac{|k|}{\sqrt{t}}\\right). $$$$ \mathbb P(\tau_k < t) = 2\cdot\Phi\left(-\frac{|k|}{\sqrt{t}}\right). $$
