= Solution
With $\epsilon_{0123}=+1$, define the <Pauli-Lubanski pseudovector> by
$$
W_\mu=\frac12\epsilon_{\mu\nu\rho\sigma}M^{\nu\rho}P^\sigma.
$$
For a translation $a^\mu$ and a <Lorentz transformation> $\Lambda=\exp(\omega)$, a left-handed <Weyl spinor> transforms as
$$
\psi'_\alpha(x')=
\left[\exp\left(-\frac i2\omega_{\mu\nu}\sigma^{\mu\nu}\right)\right]_\alpha{}^\beta
\psi_\beta(x),
\qquad x'=\Lambda x+a.
$$
Equivalently, the active field at a fixed point uses $\psi'(x)=S(\Lambda)\psi(\Lambda^{-1}(x-a))$.
The <Clebsch-Gordan decomposition> is
$$
\left(\frac12,0\right)\otimes\left(\frac12,0\right)
=(1,0)\oplus(0,0).
$$
Indeed,
$$
\psi_\alpha\chi_\beta
=\psi_{(\alpha}\chi_{\beta)}
+\frac12\epsilon_{\alpha\beta}\psi^\gamma\chi_\gamma.
$$
The symmetric spinor is the $(1,0)$ representation, while its antisymmetric part is the Lorentz scalar $(0,0)$.
In the conventions used below, the nonzero brackets involving the supercharges are
$$
\{Q_\alpha,\bar Q_{\dot\beta}\}=2\sigma^\mu_{\alpha\dot\beta}P_\mu,
\qquad
[M^{\mu\nu},Q_\alpha]=i(\sigma^{\mu\nu})_\alpha{}^\beta Q_\beta,
$$
$$
[M^{\mu\nu},\bar Q_{\dot\alpha}]
=i(\bar\sigma^{\mu\nu})_{\dot\alpha}{}^{\dot\beta}\bar Q_{\dot\beta},
\qquad
[P_\mu,Q_\alpha]=[P_\mu,\bar Q_{\dot\alpha}]=0,
$$
and $\{Q_\alpha,Q_\beta\}=\{\bar Q_{\dot\alpha},\bar Q_{\dot\beta}\}=0$. These equations, together with the <Poincare algebra>, are the four-dimensional $\mathcal N=1$ <Super-Poincaré algebra> without central charges.
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