The digital option pays $H$ at time $T$ if $S_T \geq K$ , so its option time at time $t$ is given by
$$V_t=E_t\left[e^{-r(T-t)}H 1_{\{S_T \geq K\}}\right]=e^{-r(T-t)}H* P_t(S_T \geq K)$$
The model used is Black-model, that
$$dS_t=rS_tdt+\sigma dW_t$$
or
$$S_T=S_te^{\left(r-\frac12 \sigma^2\right)(T-t)+\sigma (W_T-W_t)}{}$$
Calculate $ P_t(S_T \geq K)$
$$ P_t(S_T \geq K)=P_t(S_te^{\left(r-\frac12 \sigma^2\right)(T-t)+\sigma (W_T-W_t)}{} \geq K)=P_t(W_T-W_t \geq\frac{log\frac{K}{S_0}-\left(r-\frac12 \sigma^2\right)(T-t)}{\sigma})$$
$W_T-W_t |W_t$ is centered and normally distributed with variance $T-t$
$$P_t(W_T-W_t \geq\frac{log\frac{K}{S_0}-\left(r-\frac12 \sigma^2\right)(T-t)}{\sigma})=P(Y \geq\frac{log\frac{K}{S_t}-\left(r-\frac12 \sigma^2\right)(T-t)}{\sigma \sqrt{T-t}})$$
where $Y \sim \mathcal{N}(0,1)$
Using the symmetry of the normal distribution,
$$P(Y \geq\frac{log\frac{K}{S_t}-\left(r-\frac12 \sigma^2\right)(T-t)}{\sigma \sqrt{T-t}})=P(Y \leq -\frac{log\frac{K}{S_t}-\left(r-\frac12 \sigma^2\right)(T-t)}{\sigma \sqrt{T-t}})$$
Define $$d_2=-\frac{log\frac{K}{S_t}-\left(r-\frac12 \sigma^2\right)(T-t)}{\sigma \sqrt{T-t}}=\frac{log\frac{S_t}{K}+\left(r-\frac12 \sigma^2\right)(T-t)}{\sigma \sqrt{T-t}}$$
$$P(Y \leq -\frac{log\frac{K}{S_t}-\left(r-\frac12 \sigma^2\right)(T-t)}{\sigma \sqrt{T-t}})=P(Y \leq d_2)=N(d_2)$$
where $N$ is the cdf of a standard normal variable.
Finally,
$$V_t=E_t\left[e^{-r(T-t)}H 1_{\{S_T \geq K\}}\right]=e^{-r(T-t)}H*N(d_2)$$