Understanding circular convolution
Your first version of $J(a*b)[n]$ is correct:

$$J(a*b)[n] = \sum_{p=0}^{N-1} a[p]b[-n-p]$$

because you simply have to flip $n$, and not the summation index.

The problem is in your formulation of the circular convolution of $(Ja)[n]$ and $(Jb)[n]$. It should be

$$((Ja)*(Jb))[n]=\sum_{p=0}^{N-1} a[-p]b[-n+p]$$

because you have to negate the arguments of both sequences, so $p$ becomes $-p$, and $n-p$ becomes $-n+p$. Now you can simply change the sign of $p$ because it doesn't matter in which order you compute the sum. This gives

$$((Ja)*(Jb))[n]=\sum_{p=0}^{N-1} a[p]b[-n-p]$$

which is identical to $J(a*b)[n]$ as given above, as it should be.