It is possible to normalize this distribution, see Fractional logit model in Stan - #2 by bbbales2
\int_{0}^{1} p^{x}(1-p)^{1-x} d x=\frac{1-2 p}{2 \tanh ^{-1}(1-2 p)}
However, the parameter p doesn’t really mean anything, but the mean of the distribution is \int_{0}^{1} \frac{x p^{x}\left((1-p)^{1-x}\left(2 \tanh ^{-1}(1-2 p)\right)\right)}{1-2 p} d x=\frac{p}{2 p-1}+\frac{1}{2 \tanh ^{-1}(1-2 p)}, a function of p with no analytical functional inversion.
(picture from Wolfram Alpha)
In addition the distribution is undefined when p=0.5, but the limit is 1.
It should be possible to add a RNG with inverse transform sampling based on the inverse CDF: x=\frac{\log \left(-\frac{p-1}{2 p q-p-q+1}\right)}{\log (1-p)-\log (p)}
(CDF: \int_{0}^{x} \frac{p^{x}(1-p)^{1-x}\left(2 \tanh ^{-1}(1-2 p)\right)}{1-2 p} d x=-\frac{p^{x}(1-p)^{1-x}+p-1}{1-2 p}=q)
That being said, it is still a fairly worthless distribution.
EDIT: Fixed some derivations. Some step were incorrect, yielding the rest incorrect.
Addition: This is, as pointed out by @spinkney, also known as the continuous Bernoulli distribution.