 Solution

Choose the extended supersymmetry normalizationThe central charge in supersymmetry is antisymmetric in $A,B$, because the full anticommutator is symmetric and ${ϵ}_{\alpha \beta }$ is antisymmetric. In a massive one-particle state rest frame, $P_{\mu }=(M,0,0,0)$, and the first anticommutator becomes $2M{\delta }_{\alpha \beta }{\delta }^A{}_B$.

A unitary skew-diagonalization of an antisymmetric matrix puts $Z$ into blocks $\left(\begin{array}{cc}0& z_r\\ -z_r& 0\end{array}\right)$, with a remaining zero direction if $\mathcal{N}$ is odd. For one block write $z=|z|e^{i\phi }$, and label its two supersymmetries $1,2$. The combinationshaveFor example the mixed contribution to the first expression is $\pm (e^{-i\phi }z+e^{i\phi }\bar{z})/2=\pm |z|$. This is the phase-aligned version of taking a supercharge minus a unitary combination of its adjoints, as suggested by the hint.

For every operator $X$ and normalized state $|v⟩$,Applying this to $A_{-},B_{-}$ in every block proves the BPS boundIf the algebra is instead written with $2Z$ on its right side, the same result is $M\ge {\max }_r|Z_r|$; the factor is purely a central-charge convention.

A BPS state saturates at least one of these inequalities. The corresponding zero-norm combinations of supercharges and their adjoints annihilate the representation, so part of supersymmetry is preserved. Its massive supermultiplet is shortened: each saturated block removes two complex fermionic oscillators, giving $(2j+1)2^{2\mathcal{N}-2k}$ states from a spin-$j$ Clifford vacuum when $k$ blocks saturate, before any necessary CPT completion of a supermultiplet. It preserves $4k$ of the $4\mathcal{N}$ real supercharges. Its mass is tied to the central charges in supersymmetry and remains protected while it stays on the same short branch. Stability is not unconditional: a wall of marginal stability can permit a charged BPS state to decay into other BPS states with the same total charge.