Describing a circular current loop as delta functions

EDITED ANSWER: The delta distribution $\delta(x)$ is not unique. It is invariant under transformations of the form $\delta(x) \to f(x)\delta(x)$ where $f(0) = 1$. This is because it is really a distribution and not a function. It is mathematically improper to talk about $\delta(x)$ instead of $\int \delta(x)dx$. Derivations of the term you're interested in will not be unique either. You can show that

$$ I = I\int \delta(r-a)dr = I\int \delta(r-a)\delta(\theta - \pi/2)drd\theta = I\int\frac{\delta(r-a)}{r}\delta(\theta - \pi/2)rdrd\theta. $$

From this expression, it is apparent that we can write the current as

$$ I = \int J_{\phi}rdrd\theta \implies J_\phi = I \frac{\delta(r-a)}{r}\delta(\theta - \pi/2). $$

This result leaves out the $sin(\theta)$ and replaces $a$ with $r$. It really makes no difference because distributions are mathematically represented by $(T, F)$ where $F$ is a set of continuous functions and $T$ is the map from $F \to \mathbb{R}$. While it is common for physicists to represent distributions as just the mapping (i.e., $T = (T,F)$), this is somewhat false and leads to the non-unique representation of quantities like $\delta(x)$.

If you note that $$\delta(\cos\theta)=\frac{\delta(\theta-\pi/2)}{\sin\theta}$$ Then you can see that the sine terms actually cancel out.

See, you are required to find volume current density $J_\phi$. Though its name is volume current density, you know it is the current flowing per unit surface area. Now the subscript $\phi$ in $J_\phi$ denotes it is flowing in the $\hat{\phi}$ direction. Now in the spherical polar co-ordinate the infinitesimal length elements along the direction $\hat{r},\hat{\theta} \rm\ and\ \hat{\phi}$ are dr, $rd\theta$ and $r\sin\theta d\phi$ respectively. So $dr\times rd\theta$ is the area you are interested in.

Now looking at your figure I am interested in rewriting the current I.

$$I=I\int \delta(r-a) dr= I\int \delta(r-a) dr\ \delta\big(\cos\theta-cos(\pi/2)\big) d(cos\theta)$$ $$=I\int \delta(r-a) dr\ \delta\big(\cos\theta-0\big) d(cos\theta)$$ $$=I\int \delta(r-a) dr\ \delta\big(\cos\theta\big) d(cos\theta)$$ $$=I\int \delta(r-a) dr\ \delta\big(\cos\theta\big) \sin\theta d\theta$$ $$=I\int \delta(r-a)\delta\big(\cos\theta\big) \sin\theta dr\ d\theta$$ $$=I\int \frac{ \delta(r-a)\delta(\cos\theta) \sin\theta}{r} {dr\ r d\theta}$$

So you see your $$J_\phi = I\frac{ \delta(r-a)\delta(\cos\theta) \sin\theta}{r}$$