A free particle has constant potential energy, without loss of generality $0$, so the Minkowski square norm of its $4$-momentum is$$E^2/c^2-\vec{p}^2=m^2c^2,$$with $\vec{p}$ the $3$-momentum an $m$ the rest mass. The RHS is invariant, meaning observer-independent, do not confuse this with conserved, meaning time-independent. (I'll discuss conserved quantities later.) Now to generalize: when the particle isn't free, we generalize this equation while preserving its RHS, say as$$(E-U)^2/c^2-(\vec{p}-\vec{Q})^2=m^2c^2.$$This is the Minkowski square norm of the $4$-momentum minus a $4$-vector whose energy part is the potential energy $U$. It therefore makes sense to call the $3$-vector $\vec{Q}$ the potential momentum.
The source you quote uses that term with a very different meaning, then explains why that one doesn't make sense: in other words, it explains why a different "energy is like this, therefore momentum is like that" analogy doesn't work. The familiar kinetic momentum is conserved, rather than being one of two terms whose sum is conserved while the terms themselves separately aren't. By contrast, energy conservation is only the conservation of a sum of two non-conserved energy terms, one kinetic, one potential. So while one can store a non-conserved amount of energy as what we call potential, momentum doesn't work the same way.