= Solution
Take $x$ vertically downward and $y$ outward from the wall, with the film occupying $0<y<h$. The ambient <pressure> is
$$
p_a(x)=p_0+\rho_agx.
$$
At $y=h$, normal stress continuity gives $p=p_a$ and zero ambient shear gives $\mu U'(h)=0$; at the wall, no slip gives $U(0)=0$.
For steady parallel flow $u=(U(y),0)$, incompressibility is automatic and Navier–Stokes reduces to
$$
p_y=0,\qquad0=-p_x+\mu U''+\rho g.
$$
Since $p_x=\rho_ag$, integration yields
$$
\boxed{p=p_0+\rho_agx,\qquad
U(y)=\frac{(\rho-\rho_a)g}{\mu}\left(hy-\frac{y^2}{2}\right).}
$$
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