 Solution

Use $c=1$ and a spatially flat background dominated by the stated perfect fluid, so $4\pi G{a}^2{\rho }_0=3{\mathcal{H}}^2/2$. In Newtonian gauge, the scalar energy and momentum constraints from the linearized Einstein field equations combine to eliminate the potential's time derivative and give the stated relativistic Poisson constraint. Its source contains the velocity correction to the coordinate density perturbation. On sub-horizon scales that correction is small for the ordinary growing modes, recovering the Poisson equation for Newtonian gravity; near or outside the horizon the gravitational field cannot be sourced by $\delta$ alone. Equality of the two metric potentials also assumes negligible scalar anisotropic stress.

The fluid equations are the linear energy-continuity and momentum-Euler parts of ${\nabla }_{\mu }{T}^{\mu \nu }=0$. Here $c_s^2=\delta p/\delta \rho$ is the relevant dimensionless sound-speed squared in the adopted fluid closure; for a constant-$w$ adiabatic barotropic fluid it equals $w$. A fluid with entropy perturbations needs additional pressure information. The equations actually use $\theta =\nabla \cdot \mathbf{v}$ in Fourier space, the peculiar-velocity divergence. Calling it simply momentum density is imprecise: it is the divergence of the velocity, with momentum weighted by ${\rho }_0+p_0$ before normalization. In particular $\theta$ must have inverse-conformal-time units for the displayed continuity equation to be dimensionally consistent.