Given $\huge\frac{p(x=r)}{p(x=n-r)}$ is independent of $n$ and $r$.
$\Rightarrow \huge\frac{c^{n}_{r} p^{r}(1-p)^{n-r}}{c^{n}_{n-r} p^{(n-r)} (1-p)^{r}}$
=$\large p^{2r-n} (1-p)^{n-2r}$
$\Large\left(\frac{1-p}{p}\right)^{n-2r}$
Will be independent of $n$ and $r$ noly if $1-p=p$
$1=2p$
$or \;p=\large\frac{1}{2}$.