The geodesic equation can be written equivalently when $X$ is parametrized with proper time:
$$
\frac{d^2X^\mu}{d\tau^2} + \Gamma^\mu_{\nu\alpha}\frac{dX^\nu}{d\tau} \frac{dX^\alpha}{d\tau} = 0\,.
$$
Here, $\tau\mapsto X^\mu(\tau)$ is a world line whose four velocity is $\frac{dX^\mu}{d\tau}$ and whose acceleration is $\frac{d^2X^\mu}{d\tau^2}\,.$
Now suppose we have a four-velocity field $U^\mu$ at every point of space time (or at least around the world line $X^\mu$) such that
$$
\frac{dX^\mu(\tau)}{d\tau}=U^\mu(X^\mu(\tau))\,.
$$
By the chain rule
$$\tag{1}
\frac{d^2X^\mu(\tau)}{d\tau^2}=\frac{dX^\mu(\tau)}{d\tau}\partial_\nu U^\mu(X^\mu(\tau))=(U^\nu\partial_\nu U^\mu)(X^\mu(\tau))\,.
$$
Dropping the clumsy argument $X^\mu(\tau)$ the geodesic equation then implies
$$
U^\nu\partial_\nu U^\mu+\Gamma^\mu_{\nu\alpha}U^\nu U^\alpha=U^\nu\nabla_\nu U^\mu=0.
$$
Four momentum $P^\mu$ is just rest mass $m_0$ times four velocity $U^\mu\,.$ This does not alter the geodesic equation.
Few remarks.
In curved space time $\frac{d^2X^\mu}{d\tau^2}$ is not the four acceleration. Since by (1) it is $U^\nu\partial_\nu U^\mu$ which does not transform as a tensor due to the partial derivative there.
The correct way of defining four acceleration is
$$\tag{2}
A^\mu\equiv\frac{d^2X^\mu}{d\tau^2}+\Gamma^\mu_{\nu\alpha}\frac{dX^\nu}{d\tau}\frac{dX^\alpha}{d\tau}\,.
$$
As above this is seen to be equal to $U^\nu\nabla_\nu U^\mu\,.$
This is the covariant derivative of $U$ in the direction of $U\,.$
Some authors also write $\dot U^\mu$ for $A^\mu$
and the shortest form of the geodesic equation is
$$
\boxed{A^\mu=0\,.}
$$
