= Solution
First fix a source convention needed for physical coordinates. The PDF's displayed definition of $\sigma_z$ omits a factor of $i$: half the <commutator> of the two Hermitian transverse matrices is anti-Hermitian and cannot supply a real Bloch coordinate. The Hermitian choice consistent with the supplied dynamical matrix is
$$
\sigma_z=\frac{[\sigma_x,\sigma_y]}{2i}
=\begin{pmatrix}-1&0\\0&1\end{pmatrix}.
$$
With this convention the three Pauli expectation values use the unit-radius normalization, unlike the orthonormal coordinates of Question 2. For example, writing $K=\begin{pmatrix}0&a\\b&0\end{pmatrix}$ with $a=1+\lambda$, $b=-\lambda$, the dissipator gives rates $-(a-b)^2/2$ for $x$, $-(a+b)^2/2$ for $y$, and population equation $\dot z=b^2-a^2-(a^2+b^2)z$. The <Quantum Hamiltonian> adds $2\alpha z$ to $\dot x$ and $-2\alpha x$ to $\dot z$. This independently reproduces the supplied real Bloch dynamics.
Put $q=2\lambda+1$. Since $y$ decays independently, every <steady state> has $y_*=0$. The reduced steady-state equations are
$$
q^2x_*-4\alpha z_*=0,\qquad
4\alpha x_*+(q^2+1)z_*=-2q.
$$
Their coefficient <determinant> is
$$
D=q^2(q^2+1)+16\alpha^2.
$$
It is strictly positive unless $q=0$ and $\alpha=0$. Inverting this two-by-two system gives
$$
\boxed{x_*=-\frac{8\alpha q}{D},\qquad
z_*=-\frac{2q^3}{D}.}
$$
Thus the reduced system has a unique <steady state> except at $\alpha=0$, $\lambda=-1/2$. At that exceptional point the equations reduce to $\dot x=0$, $\dot z=-z/2$, so every $(x,0)$ is stationary; physical coordinates restrict $-1\leq x\leq1$.
As a consistency check on positivity in the nonsingular case,
$$
1-x_*^2-z_*^2=
\frac{[16\alpha^2+q^2(q^2-1)]^2}{D^2}\geq0.
$$
The calculated state is therefore in the unit Bloch disk. The original PDF correctly labels the second steady-state expression as $z_*$; the TeX conversion's repeated $x_*$ is another transcription defect, not a second formula for $x_*$.
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