= Solution
Let $N=\sqrt{K^aK_a}>0$ on the fluid region and choose the <Killing vector field> $K$ to be future-directed. The normalized <four-velocity> is
$$
\boxed{U^a=\frac{K^a}{N}.}
$$
The <Killing equation> implies both $K^b\nabla_bN=0$ and
$$
K^b\nabla_bK_a=-K^b\nabla_aK_b=-\frac12\nabla_a(K^bK_b).
$$
Consequently its <Killing-flow four-acceleration> is
$$
U^b\nabla_bU_a=\frac1{N^2}K^b\nabla_bK_a=-\nabla_a\log N.
$$
For a scalar, the <Lie derivative> is $\mathcal L_Kp=K^b\nabla_bp$. Thus the assumed stationarity of the <pressure> gives $U^b\nabla_bp=0$. These are the ingredients needed for <hydrostatic equilibrium along a Killing vector>; hypersurface orthogonality of $K$ is not needed.
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