Angular momentum and linear momentum are fundamentally different quantities. At a basic level, they are generators/charges of two very different symmetries and are thus, conserved, according to Noether theorem, under two very different symmetries. In particular, angular momenta are the generators of the rotational symmetry, i.e. $SO(d)$ in a $d-$dimensional space while the linear momenta are the generators of the translational symmetry. The two symmetry groups are absolutely distinct, to wit, one (the group of rotations) is compact while the other (the group of translations) is not. To make the distinction even sharper, in $d$ dimensions, one would have $d$ linear momenta while $\frac{d(d-1)}{2}$ angular momenta. It is just a coincidence that in $d=3$, both those expressions yield $3$ which has the potential of creating confusion that they both might be representing the same quantity.
Coming to the specific situation OP raises, one cannot ascertain the linear momentum of the particle simply from knowing its angular momentum. Again, to wit, the angular momentum is conserved in a Kepler problem (actually, something even better is conserved in a Kepler problem but I won't go there for the moment) and thus, it simply never changes throughout the motion. But, the linear momentum continuously changes as the particle orbits around the focus of its orbit.