[Kupershmidt](https://arxiv.org/abs/math/0001187) argues for 
$$\mu_{K}([j]_q)\propto(p;q)_{j-1}\,q^j,\;\;[j]_q=\frac{q^j-1}{q-1},$$ 
as the "natural" $q$-geometric distribution, because it produces the $q$-analog of the Pascal (negative binomial) distribution. It has a simple normalization,
$$\mu_{K}([j]_q)=\frac{p}{1-(p;q)_\infty}\,(p;q)_{j-1}\,q^j.$$
Since $(p;1)_n=(1-p)^n$ and $[j]_1=j$, for $q\rightarrow 1$ the classical geometric distribution $\mu(j)=p(1-p)^{j-1}$ is recovered (the $\mu_1$ from the OP with $p\mapsto 1-p$).