= Solution
Use the rotation convention of part (a), so the orientation drift is $B^{-1}(\mathbf e_3-p_3\mathbf p)-\Omega\mathbf e_2\times\mathbf p$. The steady <gyrotactic orientation Fokker-Planck equation> on the unit sphere is
$$
\nabla_p\cdot\left[\left\{\frac1B(\mathbf e_3-p_3\mathbf p)-\Omega\mathbf e_2\times\mathbf p\right\}f-D_R\nabla_pf\right]=0,\qquad \int f\,d\Omega=1.
$$
Let $\eta=(B\Omega)^{-1}$, $\lambda=(BD_R)^{-1}$, and introduce the rotation generator $\mathcal A=(\mathbf e_2\times\mathbf p)\cdot\nabla_p$. Dividing by $\Omega$ gives
$$
-\mathcal Af+\eta\nabla_p\cdot[(\mathbf e_3-p_3\mathbf p)f]-\frac\eta\lambda\Delta_pf=0.
$$
The leading isotropic density is $f_0=1/(4\pi)$. At first order, writing $f=f_0+\eta f_1+\cdots$, rotational <diffusion> makes no contribution because $\Delta_pf_0=0$. The tangential drift is the spherical gradient of $p_3$, whose divergence is $\Delta_pp_3=-2p_3$. Therefore
$$
\mathcal Af_1=-2f_0p_3.
$$
In the polar convention of the paper, $p_1=\sin\theta\cos\phi$, and $\mathcal Ap_1=p_3$. Thus, in the specified first-harmonic form,
$$
\boxed{f=\frac1{4\pi}-\frac1{2\pi B\Omega}\sin\theta\cos\phi+O((B\Omega)^{-2}),\qquad\alpha=-\frac1{2\pi}.}
$$
The correction integrates to zero, preserving normalization. The first correction has no dependence on $\lambda$ because the isotropic leading density has zero spherical Laplacian; $\lambda$ enters the next-order equations.
Spherical symmetry gives $\int p_ip_jd\Omega=4\pi\delta_{ij}/3$. Consequently the <rapid-rotation mean gyrotactic orientation> is
$$
\boxed{\langle\mathbf p\rangle=-\frac{2}{3B\Omega}\mathbf e_1+O((B\Omega)^{-2}).}
$$
It is this small horizontal swimming bias, rather than a mean vertical orientation at leading order, that enters the concentration equation of part (c).
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