Why is the energy-momentum tensor $T_{\mu \nu}$ the source of the gravitational field instead of the momentum vector $P_{\mu}$?

Because for a continuous distribution of matter (or energy), such as a fluid, what you really want is the density of momentum, and this density is given by the energy-momentum tensor.

If you have a fluid (or, again, any distribution of energy), you can define at each point the velocity $\mathbf{u}$, which in spacetime generalizes to the four-velocity $u^\mu$. No problem so far. But if you want the momentum at each point, that's the mass times the velocity. And we run into a problem, because the mass of a single point is zero! Mass is not a quantity that can be defined at each point; it's not a field.

That's fine, because we know that we really should have used the mass density. And so in Newtonian mechanics we can write the momentum density as $\rho \mathbf{u}$, and the energy density as $\frac12 \rho u^2$. But in relativity this doesn't really work, because density is not a scalar: it changes after a Lorentz boost. Some math reveals that the energy density transforms as the zero-zero component of a tensor (picking up two factors of $1/\sqrt{1-v^2}$), which is the energy-momentum tensor.

Or to put it another way: the energy-momentum density, whatever it is, must have two indices. You need one index to tell me what your time axis is; and then I can take the 3D space orthogonal to that time axis, and give the momentum density in that space, accounting for the other index. But I cannot define a density if you don't tell me how you choose to split spacetime into space and time.

Here is one line of reasoning: Recall that a source $J_k$ is (minus) the derivative of the effective action $\Gamma[\phi_{\rm cl}]$ wrt. the classical field $\phi^k_{\rm cl}$. By analogy, one may consider the Hilbert/metric SEM tensor $$T^{\mu\nu}~:=\pm\frac{2}{\sqrt{|g|}}\frac{\delta S}{\delta g_{\mu\nu}}$$ as the source of the gravitational field $g_{\mu\nu}$, cf. OP's title question.

Actually we do use something a bit like $P_\mu P_\nu $. If you sum the expression $P_\mu v_\nu $ for all the particles of matter in a region, you get $ T_{\mu\nu} $, at least for the matter element. You still need the e.m. component from photons, but you can see that it must be included because of conservation of energy-momentum.

Einstein had realised much earlier that curvature must be involved, by thinking of the twin paradox which lead him to express the equivalence principle. One cannot use Riemann because that would mean spacetime was flat in regions with no mass. He tried using Ricci, but it did not work. Only when he found that the Einstein tensor obeys the contracted Bianci identity, and consequently automatically conserves energy-momentum did he find the correct equation.