= Solution
Use the same four Grassmann generators and a fixed order $a,b,c,d$. Their squares vanish, so
$$
(bc-ad)^2=-bcad-adbc=-2abcd.
$$
Meanwhile the two self-contractions are
$$
(\psi\psi)=dc-cd=-2cd,\qquad
(\chi\chi)=ba-ab=-2ab,
$$
and moving the pair $cd$ past $ab$ gives no sign, yielding $(\psi\psi)(\chi\chi)=4abcd$. Thus the <square of a Weyl spinor bilinear> is
$$
\boxed{(\chi\psi)^2=-\frac12(\psi\psi)(\chi\chi),\qquad D=-\frac12.}
$$
The order of individual odd components, rather than an informal commuting-spinor substitution, fixes this sign. For commuting spinors both self-contractions vanish, so no corresponding universal factorization of the generally nonzero mixed square is possible.
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