Here is a problem from probability theory textbook by Larson.
Fifty-five percent of the registered voters in Sheridanville favor their incumbent mayor in her bid for re-election. If four hundred voters go to the polls, approximate the probability that the race ends in a tie.
We are looking for the probability of a tie, so we let $X$ be the number of voters going for the incumbent. We want the probability that $P(X=200)$ since there are 400 voters.
We know that $n=400$ and $p=0.55$. We can continue and say that we are essentially looking for $P\left (199.5 \leq X \leq 200.5\right)$. We can make the transformation of $X$ using the theorem: $\dfrac{X-np}{\sqrt{npq}}$. The same things can be done to the left and right bounds 199.5 and 200.5.
My question is: I don't know which $p$ to use. Here, $p=0.55$ because I let my $X$ is the number voting the incumbent. But, if we let $X$ be the number voting for the challenger, then surely $p=0.45$. The numerator changes but not the denominator, so I am confused about which $X$ is the correct representation?
You'll get the same answer either way. If $X$ denotes the number of votes going to the incumbent, then $X$ has binomial $(n=400, p=.55)$ distribution. If $Y$ denotes the number of votes going to the challenger, then $Y$ has binomial $(n=400, p=.45)$ distribution. But $P(X=200) = P(Y=200)$. You know this is true without calculation, because the event $\{X=200\}$ is the same as the event $\{Y=200\}$.
As for your confusion about the numerator changing but not the denominator, note that standardization of $X$ leads to a probability of the form $P(a\le Z\le b)$. You can check that standardization of $Y$ leads to a probability of the form $P(-b \le Z\le -a)$. These probabilities are equal, since the standard normal distribution is symmetric about zero.