As said in comments, maybe just the Mahalanobis distance could work. But, as explained in Intuition of the Bhattacharya Coefficient and the Bhattacharya distance? and other places, often an alternative which might work better is the Bhattacharyya distance based on the Bhattacharyya coefficient
$$
BC(p,q) = \sum_{k=0}^n \sqrt{p(k) q(k)}
$$
here written in the discrete case relevant for the binomial distribution. Now let $p, q$ respectively be binomial distributions with the same index $n$ and probability parameters $p,q$. Then the Bhattacharyya coefficient becomes
$$
BC(p,q) = \sum_{k=0}^n \binom{n}{k} (pq)^{k/2} ((1-p)(1-q))^{(n-k)/2}
$$
which can be simplified to
$$
BC(p,q) = (\frac{\widetilde{pq}}{\widetilde{(1-p)(1-q)}}+1)^n \widetilde{(1-p)(1-q)}^n
$$
where the symbols with a tilde over them represents the geometric mean of $p,q$ and of $(1-p),(1-q)$, respectively. Then the Bhattacharrya distance is given by
$$
BD(p,q)= -\ln BC(p,q)
$$
You can easily check that if $p=q$ then this distance becomes zero.
Below is a plot of this distance function for the case $n=10$:

and for the record the R code used to produce it:
BD <- Vectorize( function(p, q, n) {
PQ <- sqrt(p*q)
PQC <- sqrt((1-p)*(1-q))
-n*log((PQ/PQC)+1) - n*log(PQC)
}
)
x <- y <- seq(0.01, 0.99, by=0.01)
z <- outer(x, y, FUN=function(x, y) BD(x, y, 10))
levels <- c(seq(0.1, 1.5, by=0.2), 2.5, 3.5, 4.5, 5.5, 6.5, 7.5, 8.5, 9.5, 10.5)
image(x, y, z, col=terrain.colors(15), xlab="p", ylab="q" )
title("Bhattacharyya distance between binomial n=10")
contour(x, y, z, levels=levels, add=TRUE)