= Solution
Work in the dilute, single-scattering limit and add the diffuse scattered <light intensities> of randomly positioned particles. The incident <unpolarized light> has equal, mutually uncorrelated $x$ and $y$ electric components. Each particle therefore has an induced <electric dipole moment> along its permitted axis, with the same mean squared magnitude proportional to $|\alpha|^2$.
For an observer in direction $\widehat{\mathbf n}=(\sin\psi,0,\cos\psi)$, choose transverse unit directions $\mathbf e_\parallel=(\cos\psi,0,-\sin\psi)$ in the scattering plane and $\mathbf e_\perp=\widehat{\mathbf y}$ perpendicular to it. The <electric dipole radiation> field is proportional to the transverse projection of the <electric dipole>. Hence an $x$ <electric dipole> contributes a <wave amplitude> factor $\cos\psi$ along $\mathbf e_\parallel$, while a $y$ <electric dipole> contributes its full <wave amplitude> along $\mathbf e_\perp$.
Let $C>0$ collect the common incident <light intensity>, <electric polarizability>, distance and <frequency> factors. Then
$$
I_\parallel=CN_1\cos^2\psi,\qquad I_\perp=CN_2.
$$
The two components are uncorrelated, so the <coherency matrix> is diagonal in this basis. The <degree of polarization> is
$$
\boxed{\mathcal P_{xz}(\psi)=\frac{|I_\perp-I_\parallel|}{I_\perp+I_\parallel}
=\frac{|N_2-N_1\cos^2\psi|}{N_2+N_1\cos^2\psi}.}
$$
A signed degree of <linear polarization>, defined positive perpendicular to the scattering plane, has the same expression without the absolute value. The dominant direction is $\widehat{\mathbf y}$ when $N_2>N_1\cos^2\psi$, and $\mathbf e_\parallel$ when the inequality reverses. If the denominator vanishes there is no scattered signal and its <degree of polarization> is undefined. A coherent exactly forward component, including the transmitted incident beam, is not part of this diffuse-intensity calculation; random particle positions do not by themselves destroy forward <optical coherence>.
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