You can use such an approximation but there are known analytical prices. You have a special case in which the stock price is normally distributed. See Bachelier Model.
Set $\mu=r-q$ (if you have dividends, or simply $\mu=r$ if there are no dividends). So if you change from the real worl probability measure $\mathbb{P}$ to the risk-neutral measure $\mathbb{Q}$ you get that $\mathrm{d}S_t=(r-q)\mathrm{d}t+\sigma \mathrm{d}W_t$. Then, using risk-neutral pricing, the inital value of your claim is given by
\begin{align*}
V_0 &= e^{-rT} \mathbb{E}^\mathbb{Q}[{1}_{\{S_T< K\}}] \\
&= e^{-rT} \mathbb{Q}[\{S_T< K\}].
\end{align*}
Thus, all you need to do is to find the probability distribution of $S_T$ under $\mathbb{Q}$. Using again that $\mathrm{d}S_t=(r-q)\mathrm{d}t+\sigma \mathrm{d}W_t$, we see that $(S_t)$ is an arithmetric Brownian motion under $\mathbb{Q}$ and thus normally distributed. Furthermore,
\begin{align*}
S_T= S_0+(r-q)T + \sigma W_T \sim N\big(S_0+(r-q)T,\sigma^2T\big),
\end{align*}
since $W_T\sim N(0,T)$. Now, set $m=S_0+(r-q)T$ and $s=\sigma\sqrt{T}$. Then, $S_T=m+sZ$ where $Z\sim N(0,1)$. Thus,
\begin{align*}
V_0 &= e^{-rT} \mathbb{Q}[\{m+sZ< K\}]\\
&= e^{-rT} \mathbb{Q}\left[\left\{Z< \frac{K-m}{s}\right\}\right]\\
&= e^{-rT} \Phi\left(\frac{K-m}{s}\right)\\
&= e^{-rT} \Phi\left(-\frac{S_0-K+(r-q)T}{\sigma\sqrt{T}}\right) \\
&= e^{-rT} \left(1- \Phi\left(\frac{S_0-K+(r-q)T}{\sigma\sqrt{T}}\right)\right),
\end{align*}
where $\Phi$ denotes the cumulative distribution function of a standard normal distribution.
Let me highlight that, of course, you can price such a claim with Euler Maruyama. You can also employ finite differences or Fourier transforms. You could even build a (binomial) tree. But if there is a simple analytical answer available, it is to be preferred.
By the way, in the Black-Scholes model, the price of a Cash-Or-Nothing option is given by $e^{-rT}\Phi(-d_2)=e^{-rT}\big(1-\Phi(d_2)\big)$, see here.
