This question is motivated by MSE question 4199364: Bachelier model option pricing.
There, one considers the price of a stock depending on time $t$, given by the family of random variables $(S_t)_{t\in[0,\infty[}$ as
$$S_t = 1+\mu t+W_t$$ for some $\mu\in\mathbb R$, where $(W_t)_{t\in[0,\infty[}$ is a standard Wiener process. (Note that the precise form of $S_t$ is not important for the question asked below, for instance, one might ask the same question if $S_t$ is a, say, geometric Brownian motion, given by $S_t=S_0 \exp((\mu-\sigma^2/2)t+\sigma W_t)$ for constants $\mu\in\mathbb R$, $\sigma \in \mathbb{R}^+$ (i.e. the solution to $\mathrm dS_t = \mu S_t\,\mathrm dt + \sigma S_t\,\mathrm dW_t$, where stochastic differential notation was used), or more generally, any stochastic process with drift.)
The question was what the price of an option paying $1$ dollar (or any other currency) if $S_1>1$ and paying $0$ dollars if $S_1<1$ would be in an idealized market (as far as I know, the usual "idealized" market is characterized fully by some technical assumptions about which I know nothing and the condition that there is no arbitrage).
Now, naïvely I would expect the price of the option described above to be just $\mathsf P(S_1>1)$ dollars (where $\mathsf P$ is the probability measure of the probability space on which $(W_t)_{t\in[0,\infty[}$ was defined (sometimes called a "Wiener space")).
However, I was informed that, in fact, the price of the option is independent of $\mu$. Why is that so? Why does a higher probability of getting $1$ dollar not result in a higher price for the option?
Hedging and replication warm-up
Let's begin with a warm-up game.
Suppose we are playing roulette, where if we bet $\$1$ on red and the roulette lands on that colour, then we get $\$2$, otherwise we get $\$0$. At this casino, however, we know that the roulette is biased so that $P(\text{Red}) = 0.7$ and $P(\text{Black}) = 0.3$. This funky casino also has a bookie who offers to sell you a lottery ticket which pays $\$100$ if this biased roulette lands on red and zero otherwise; the bookie is selling this ticket for $\$60$. Is this ticket cheap or expensive?
In the above example, the real-world probabilities did not affect the fair-value price of the ticket and the reason for this is the law of one price: two "portfolios" with the same payoff must have the price. This law gives rise to the principle of replication. The same principle holds true for financial derivatives.
One-period binomial market model
Let's consider the simplest possible market model: one stock $(S_t)$, one bond $(B_t)$, and times $t =0, 1$. The sample space is $\Omega = \{ H, T \}$, the risk free interest rate is $r \geq 0$ and at $t = 1$, the stock is either $S_1(H) = uS_0$ (with probability $p\in[0,1]$) or $S_1(T) = d S_0$ (with probability $1-p$), where $u$ and $d$ are known, and we have a European derivative $V = g(S)$ that pays $V_1(H)$ or $V_1(T)$ at time $1$. We assume that $0 \leq d < 1+ r < u$ (exercise: why?). We wish to determine the time-zero price of $V$, call it $V_0$.
If we begin with initial wealth $X_0$ and buy $\Delta_0$ shares of the stock at time zero, we are left with $X_0 - \Delta_0 S_0$, which we invest in the bond. At time $t=1$, the value of our portfolio is $$X_1 = \Delta_0 S_1 + (1+r)(X_0 - \Delta_0 S_0) = (1+r)X_0 + \Delta_0 (S_1 - (1+r)S_0)$$ We wish to find $X_0$ and $\Delta_0$ such that $X_1(H) = V_1(H)$ and $X_1(T) = V_1(T)$; we do this by matching the discounted value at time zero of the stock+bond portfolio with the discounted value of the option payoff:
$$\begin{align*} \begin{cases} X_0 + \Delta_0 \left( \frac{1}{1+r}S_1(H) - S_0 \right) = \frac{1}{1+r} V_1(H) \\ X_0 + \Delta_0 \left( \frac{1}{1+r}S_1(T) - S_0 \right) = \frac{1}{1+r} V_1(T) \end{cases} \end{align*}$$
I leave it to you as an exercise to verify that $$\begin{align*} \Delta_0 &= \frac{V_1(H)- V_1(T)}{S_1(H) - S_1(T)} \\ X_0 &= \frac{1}{1+r} \left( \tilde{p}S_1(H) + (1-\tilde{p})S_1(T) \right) \end{align*}$$ where $\tilde{p} = \frac{1+r - d}{u-d}$. By the law of one price we argue that $V_0$ should be the cost of setting up the replicating porfolio, so that $V_0$ is equal to $X_0$ found above. Note, in particular, that $X_0$ does not depend on $p$.
A continuous time model
Assume now a more realistic (yet idealised) situation where we have a market with a bond that gains compound interest $$ \mathrm dB_t = rB_t \mathrm dt$$ and a stock that follows a geometric Brownian motion $$\mathrm dS_t = \mu S_t \mathrm dt + \sigma S_t \mathrm dW_t$$ (The case for arithmetic Brownian motion is answered in the question you linked to). We have a European option that pays $V_T = h(S_T)$ at the maturity time $t=T$. E.g. for a European call option $h(S_T) = (S_T - K)^+$ or for the digital option you described above, $h(S_T) = \chi_{\{S_T > 1\}}$. We wish to find the price of this option for $t < T$, i.e. $V_t = f(t, S_t)$ (we can and will assume that $f \in C^2$).
To that end, we set up a replicating portfolio in the stock and bond $\Pi_t = a_t S_t + b_t B_t$. We require that this portfolio satisfy the self-financing condition: $$\mathrm d \Pi_t = a_t\mathrm dS_t + b_t\mathrm dB_t$$ which amounts to asking that during $0 < t < T$ there are no cash inflows or outflows. If we are to match the price of the option with the value of our portfolio we must demand that $\mathrm dV_t =\mathrm d\Pi_t$. By Itô's lemma: $$\mathrm dV_t =\mathrm df(t,S_t) = \left( f_t (t,S_t) + \frac{1}{2} \sigma^2 S_t^2 f_{xx}(t,S_t) + f_x(t,S_t)\mu S_t \right)\mathrm dt + f_x(t,S_t)\sigma S_t\mathrm dW_t. $$
Substituting the stock model in our dynamics for $\Pi_t$, we get: $$\mathrm d \Pi_t = (a_t \mu S_t + b_t r B_t)\mathrm dt + a_t \sigma S_t\mathrm dW_t$$
Matching the diffusion terms, we observe that our portfolio must hold $a_t = f_x(t,S_t)$ to hedge away the randomness in the stock. We call this delta-hedging. Using the value of our portfolio, this implies that $b_t B_t = \Pi_t - f_x(t,S_t) S_t$. Finally, matching the drift terms in $\mathrm dV_t$ and $\mathrm d\Pi_t$, we get: $$f_t (t,S_t) + \frac{1}{2} \sigma^2 S_t^2 f_{xx}(t,S_t) + f_x(t,S_t)\mu S_t = f_x(t,S_t) \mu S_t + r (\Pi_t - f_x(t,S_t) S_t).$$ By cancelling $f_x(t,S_t) \mu S_t$ on both sides and after replicating $V_t = \Pi_t = f(t,S_t)$, we obtain the famous Black-Scholes PDE:
$$f_t (t,S_t) + \frac{1}{2} \sigma^2 S_t^2 f_{xx}(t,S_t) = r (f(t,S_t) - f_x(t,S_t) S_t)$$
Which has the terminal boundary condition $f(T,S_T) = h(S_T)$. Of note, the PDE does not depend on $\mu$, but it does depend on $r$.
Final remarks
Mechanically, we have been able to show that delta-hedging allows us to cancel away $\mu$ in our computation of the option price. Perhaps the one-period model can serve as intuition for why this happens: you may extend this to a multi-period model and then approximate the continuous time model arbitrarily by the multi-period model.
The answer I posted in the question you linked uses a different argument, which gets rid of $\mu$ via a martingale argument and Girsanov's theorem. This idea is alluded to in the link shared by @lulu, and you should consult it for a further and richer understanding for why $\mu$ vanishes away in our pricing formula.
Finally, you can't ask for $S_t$ to be any "stochastic process with drift". A consequence of the Fundamental Theorem of Asset Pricing (see e.g. [1]) implies that the physical model for the stock price must be a semi-martingale, so something like a geometric fractional Brownian motion would not be an admissible stock model and this argument would fail for it (see e.g. page 170 in [2]).
[1] Delbaen, Freddy, and Walter Schachermayer. "A general version of the fundamental theorem of asset pricing." Mathematische annalen 300.1 (1994): 463-520.
[2] Biagini, F., Hu, Y., Øksendal, B., & Zhang, T. (2008). Stochastic calculus for fractional Brownian motion and applications. Springer Science & Business Media.