= Solution
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A stationary viscous wall imposes the <no-slip boundary condition>, $u=v=0$, whereas the outer <inviscid flow> has $u=Ax$. At large <Reynolds number>, the transition is confined to a thin <viscous boundary layer>, where normal <velocity gradients> amplify the otherwise small viscous term.
Take a horizontal scale $L$, an outer speed $U=AL$, and a <boundary layer thickness> $\delta\ll L$. The <continuity equation> gives $v=O(U\delta/L)$. Write $\epsilon=\delta/L$ and $\mathrm{Re}_L=UL/\nu$. The tangential <Navier-Stokes equation> has scales
$$
u u_x+v u_y=O(U^2/L),\qquad
\nu u_{yy}=O(\nu U/\delta^2),\qquad
\frac{\nu u_{xx}}{\nu u_{yy}}=O(\epsilon^2).
$$
Balancing <convective acceleration> with normal viscous diffusion gives $\epsilon=O(\mathrm{Re}_L^{-1/2})$. The leading tangential equation and <incompressible flow> constraint are
$$
\boxed{u u_x+v u_y=-\frac1\rho\frac{\partial p_\infty}{\partial x}+\nu u_{yy},\qquad
u_x+v_y=0.}
$$
Here and below $\nu$ is the <kinematic viscosity>, while $u$ is the horizontal <velocity>.
For the normal equation, first remove the <hydrostatic pressure> by writing $P=p+\rho gy$. Normal <inertia> and normal viscous terms are $O(\rho U^2\delta/L^2)$, so the change of $P$ across the layer is only $O(\rho U^2\epsilon^2)$. Matching to the outer <pressure> therefore makes the leading $O(\rho U^2)$ excess <pressure> independent of $y$:
$$
P(x,y)=P_e(x)+O(\rho U^2\epsilon^2),\qquad
p_y=-\rho g+O(\rho U^2\delta/L^2).
$$
The outer expression similarly has $(p_\infty)_y=-\rho g-\rho A^2y$. Thus the printed condition
$$
\boxed{p_y=(p_\infty)_y}
$$
is a \b[leading-order <boundary layer> matching statement], with the smaller normal dynamic terms discarded on both sides. It is not an exact identity at the next order. In particular, retaining the outer $-\rho A^2y$ while dropping normal <inertia> and normal viscous terms inconsistently mixes orders. The tangential driving term is unambiguous: $-(p_\infty)_x/\rho=A^2x$.
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