= Solution
Use the usual normalization of the <Super-Poincaré algebra>, with $\sigma^\mu=(\mathbf1,\boldsymbol\sigma)$ and $\bar Q_{\dot\alpha}=Q_\alpha^\dagger$ under the corresponding index convention:
$$
\boxed{\{Q_\alpha,\bar Q_{\dot\beta}\}
=2\sigma^\mu_{\alpha\dot\beta}P_\mu.}
$$
The <supercharges> are odd operators: they turn bosonic states into fermionic states and vice versa. If $\Pi=(-1)^F$ is <fermion parity>, this statement is $\Pi Q_\alpha\Pi^{-1}=-Q_\alpha$, so
$$
\boxed{\{(-1)^F,Q_\alpha\}=0.}
$$
The same relation holds for the conjugate <supercharges>.
For a finite-dimensional physical <supermultiplet> at fixed four-momentum with energy $E>0$, let $n_B,n_F$ count physical bosonic and fermionic states. Cyclicity of the ordinary trace and the parity anticommutation imply
$$
\operatorname{Tr}\bigl(\Pi\{Q_\alpha,Q_\alpha^\dagger\}\bigr)=0:
\quad
\operatorname{Tr}(\Pi Q_\alpha^\dagger Q_\alpha)
=\operatorname{Tr}(Q_\alpha\Pi Q_\alpha^\dagger)
=-\operatorname{Tr}(\Pi Q_\alpha Q_\alpha^\dagger).
$$
Summing over the two spinor indices gives
$$
0=4E\operatorname{Tr}\Pi=4E(n_B-n_F),
\qquad\boxed{n_B=n_F.}
$$
This <supertrace pairing at positive energy> proves <boson-fermion degeneracy in a supermultiplet> for massive as well as massless positive-energy representations. It counts on-shell polarizations, not merely the names of fields. The requirement $E>0$ matters: a zero-energy <supersymmetric vacuum> can be a bosonic singlet without a paired fermionic vacuum.
If supersymmetry-breaking operators are explicitly added to the <Lagrangian>, the original <supercharges> generally no longer commute with the full Hamiltonian. They are not conserved symmetries generating finite fixed-energy physical <supermultiplets>; their original anticommutator does not equal the full translation generator with the breaking terms included. Thus the step replacing the parity-weighted anticommutator by $4E\Pi$ on a closed physical representation fails. The odd parity relation alone does not force energy degeneracy or an equal number of physical states at each mass.
This is <explicit versus spontaneous supersymmetry breaking>. In spontaneous breaking the action still has conserved <supercharges>, but the vacuum is not annihilated by them. Acting on particle excitations about that vacuum involves the broken-vacuum/Goldstino sector, so an ordinary finite particle multiplet above an invariant vacuum is no longer the correct pairing argument. The vacuum-energy statements below refer to an exact globally supersymmetric Hamiltonian, including the spontaneously broken case; they are not positivity claims for an arbitrary explicitly broken Hamiltonian.
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