Correct expression for photon 4-momentum Using that the photon is on a null geodesic, we can write the geodesic equation: 
$$g^{\mu\nu}\dot{x}_\mu\dot{x}_\nu = 0,$$
and probably also $$g^{\mu\nu}{p}_\mu{p}_\nu = 0$$.
My question: in this resource, they claim in Eq. (1) that $$p^\mu = \frac{dx^\mu}{ds} \tag{1},$$
where $s$ is an affine parameter $\neq$ proper time. Is the above definition of photon 4-momentum correct? Since the photon follows null paths I have previously understood that there is no good way of defining its 4-velocity, which is normally written as $dx^\mu/ds$.
 A: Any solution to the geodesic equation
$$
\frac{\mathrm d^{2}x^{\mu}}{\mathrm ds^{2}}+\Gamma_{\rho\sigma}^{\mu}\frac{\mathrm dx^{\rho}}{\mathrm ds}\frac{\mathrm dx^{\sigma}}{\mathrm ds}=0.
$$
will be affinely parametrized. This parametrization is only unique up to transformations $s' = a\cdot s + b$, each of which yields a different magnitude of the tangent vector $\mathrm dx^\mu / \mathrm ds$. This magnitude will remain constant along the curve as we're dealing with a metric connection.
For time-like geodesic, two of these tangent vectors (or more precisely, vector fields along the curve) are of particular interest, the so-called 4-velocity normalized to unity, and the momentum vector, whose contraction with the 4-velocity yields the particle's energy in its rest frame (aka its mass).
For light-like geodesics, there's no way to uniquely define its 4-velocity as all possible choices have the same magnitude 0, which also precludes the definition of a rest energy. However, we can still single out a momentum vector, namely the one that yields the correct energy in another frame (eg the photon energy in the frame of emitter or absorber).
As to your question, personally I would not understand the relation
$$
p^\mu = \frac{\mathrm dx^\mu}{\mathrm ds}
$$
as the definition of momentum, but rather the defintion of a certain affine parametrization $s$ compatible with the momentum.
