= Solution
Take the field along $z$ and write $x=\mu_BB$, with <Bohr magneton> $\mu_B$. The printed <Hamiltonian (quantum mechanics)> uses dimensionless angular-momentum operators, with <eigenvalues> $L(L+1)$ and $m_L$; physical angular <momenta> carry an additional $\hbar$. Thus $\beta$ and $x$ are <energies>. Assume the usual positive <spin-orbit interaction> coefficient $\beta>0$. Reversing its sign reverses the zero-field ordering but not the state count.
An <electron> has $S=1/2$. <addition of angular momentum> gives
$$
\boxed{J=L+\frac12,\quad L-\frac12\qquad(L>0).}
$$
The <coupled angular-momentum state count> can be checked directly. The product <basis> contains $(2L+1)\times2=4L+2$ states. The coupled <basis> contains
$$
[2(L+\tfrac12)+1]+[2(L-\tfrac12)+1]=(2L+2)+2L=\boxed{4L+2}.
$$
Coupling therefore reorganizes the states rather than creating or losing them. There is no hyperfine multiplicity because $I=0$.
Use $\mathbf J=\mathbf L+\mathbf S$ to obtain
$$
\mathbf L\cdot\mathbf S
=\frac12(\mathbf J^2-\mathbf L^2-\mathbf S^2),\qquad
E_{\rm SO}(J)=\frac{\beta}{2}[J(J+1)-L(L+1)-3/4].
$$
The two <spin-orbit interaction> <energies> are $\beta L/2$ and $-\beta(L+1)/2$. In a weak field, project the <Zeeman effect> onto a fixed $J$ multiplet:
$$
P_JH_ZP_J=xg_JJ_z,\qquad
\Delta E_Z=xg_Jm_J,
$$
$$
g_{L+1/2}=1+\frac1{2L+1},\qquad
g_{L-1/2}=1-\frac1{2L+1}.
$$
This is the first-order <energy> formula, not a statement that $H_Z$ has no <matrix elements> between different $J$ values. For a common $m=m_J$ and a conventional choice of relative phase, its complete coupled-basis block is
$$
H_Z=x\begin{pmatrix}
g_{L+1/2}m&-\dfrac{\sqrt{(L+1/2)^2-m^2}}{2L+1}\\
-\dfrac{\sqrt{(L+1/2)^2-m^2}}{2L+1}&g_{L-1/2}m
\end{pmatrix}.
$$
The <inter-multiplet Zeeman coupling> terms give second-order weak-field corrections because the two <spin-orbit interaction> multiplets are already separated.
In the strong-field <basis> the Zeeman part is diagonal, $E_Z=x(m_L+2m_S)$. Write
$$
H_{\rm SO}=\beta\left[L_zS_z+\frac12(L_+S_-+L_-S_+)\right].
$$
Its diagonal <matrix elements> are $\beta m_Lm_S$. Its off-diagonal <matrix elements> connect $(m_L,+1/2)$ to $(m_L+1,-1/2)$ with coefficient
$$
\frac{\beta}{2}\sqrt{L(L+1)-m_L(m_L+1)},
$$
and the reverse <spin> change has $m_L+1$ replaced by $m_L-1$ in the corresponding ladder coefficient. These terms preserve $m_J$ but mix the two product states. To first order in $\beta/x$, the <energy> correction is simply $\beta m_Lm_S$.
For the P state, $L=1$. Measured relative to its gross-structure centroid, the six low-field levels are
$$
\boxed{\begin{aligned}
E_{3/2,m}&=\frac{\beta}{2}+\frac43xm,\qquad m=\pm\frac32,\ \pm\frac12,\\
E_{1/2,m}&=-\beta+\frac23xm,\qquad m=\pm\frac12.
\end{aligned}}
$$
Thus the upper quartet starts at $\beta/2$ with slopes $+2,+2/3,-2/3,-2$ in a plot against $x$; the lower doublet starts at $-\beta$ with slopes $\pm1/3$. Their zero-field separation is $3\beta/2$.
At high field all six product-state <energies>, retaining the term linear in $\beta$, are
$$
\begin{array}{c|c|c}
m_L&m_S&E_P-E_P^{(0)}\\ \hline
-1&-1/2&-2x+\beta/2\\
-1&+1/2&-\beta/2\\
0&-1/2&-x\\
0&+1/2&+x\\
+1&-1/2&-\beta/2\\
+1&+1/2&+2x+\beta/2
\end{array}
$$
The two zero-slope levels have different $m_J$ and no mutual <matrix element>, so their first-order degeneracy does not require an additional degenerate perturbation. At finite field they split at order $\beta^2/x$.
An exact interpolation makes the requested whole-field sketch unambiguous. The stretched levels are
$$
E_{\pm3/2}=\frac{\beta}{2}\pm2x.
$$
For $m=\pm1/2$, use product states $|m-1/2,+1/2\rangle$ and $|m+1/2,-1/2\rangle$. Their <Hamiltonian (quantum mechanics)> block is
$$
H_m=\left(mx-\frac{\beta}{4}\right)\mathbf1+
\frac{x+\beta m}{2}\sigma_z+\frac{\beta}{\sqrt2}\sigma_x,
$$
so
$$
\boxed{E_m^\pm=mx-\frac{\beta}{4}\pm\frac12
\sqrt{x^2+2\beta mx+\frac94\beta^2}.}
$$
These <energies> reduce to the quartet and doublet at zero field and to the six strong-field expressions above. Within a fixed $m_J$, the nonzero off-diagonal element prevents a crossing of the two branches. Levels in different $m_J$ sectors can cross without hybridizing. For the S state, $L=0$ and the <spin-orbit interaction> term vanishes, leaving
$$
\boxed{E_{S,\pm1/2}=E_S^{(0)}\pm x}
$$
throughout this <Hamiltonian (quantum mechanics)>'s range of validity.
\Image[/past-exam-of-the-atomic-and-optical-physics-course-of-the-university-of-cambridge/2016/paper-1-zeeman-crossover.png]
{title=Weak-field multiplets evolve into the Paschen-Back product-state pattern}
{description=Original exact diagonalization in units of $\beta$. P and S <energies> use separate gross-structure references because their separation is not specified. Dashed high-field lines retain the correction linear in $\beta$; labels identify the asymptotic product states.}
For <electric dipole transitions> at low field, $\Delta J=0,\pm1$ and $\Delta m_J=0,\pm1$ apply, with the usual exclusion of $J=0\to0$. Each P state has opposite <parity> to the S state. The $P_{1/2}\to S_{1/2}$ transition has four allowed sublevel pairs. For $P_{3/2}\to S_{1/2}$, the two stretched upper states each have one destination and the other two each have two, giving six. Hence there are \b[10 allowed state-to-state channels at weak nonzero field]. Relative to the two zero-field transition <energies>, their Zeeman shifts are
$$
P_{1/2}:\quad x\left\{-\frac43,-\frac23,\frac23,\frac43\right\},\qquad
P_{3/2}:\quad x\left\{-\frac53,-1,-\frac13,\frac13,1,\frac53\right\}.
$$
In the decoupled high-field limit, an electric dipole acts on orbital coordinates, so $\Delta m_S=0$. Each of the six P product states connects to the S state with the same <spin> and $m_L'=0$. Thus there are \b[6 allowed channels in the limiting product-state description]. Their <photon> <energies> are
$$
\hbar\omega=\Delta_0+xm_L+\beta m_Lm_S,\qquad
\Delta_0=E_P^{(0)}-E_S^{(0)}.
$$
This <Paschen-Back transition-channel count> is not the count of distinct frequencies: to first order in $\beta$ the two $m_L=0$ channels coincide, giving \b[5 distinct frequencies], with shifts $0,\ \pm(x+\beta/2),\ \pm(x-\beta/2)$. If the residual <spin-orbit interaction> splitting is unresolved there are the three usual orbital Zeeman groups.
The apparent loss of four channels is a change in <eigenstate> mixing, not a change in Hilbert-space <dimension>. At finite large field, four <weak spin-forbidden Paschen-Back lines> arise from small opposite-spin admixtures, with intensities vanishing as $(\beta/x)^2$. A generic finite-field calculation therefore retains ten nonzero channels; six carry finite strength in the complete-decoupling limit. Observation direction and polarization select which of these channels a detector actually receives.
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