= Solution
Here <L-superfield> means a left-<chiral superfield>: $\bar D_{\dot\alpha}\Phi=0$, and its <complex conjugate> is an <antichiral superfield>, $D_\alpha\Phi^*=0$. Interpret the two barred derivatives as the contracted scalar $\bar D^2=\bar D_{\dot\alpha}\bar D^{\dot\alpha}$. The key point is <chiral projection by squared supercovariant derivatives>. Since the two components of $\bar D$ anticommute, any product of three barred derivatives repeats a component and is zero. Thus
$$
\bar D_{\dot\alpha}\bar D^2X=0
$$
for every <superfield> $X$, even when $X$ is antichiral. In particular, $\bar D^2\Phi^*$ is chiral.
The right side is chiral too. Because $\Phi$ is even, the <graded derivation> rule gives $\bar D_{\dot\alpha}(\Phi^2)=(\bar D_{\dot\alpha}\Phi)\Phi+\Phi(\bar D_{\dot\alpha}\Phi)=0$. Constant $m$ and $\lambda$ do not contribute derivatives. Consequently
$$
\boxed{\bar D_{\dot\alpha}(\bar D^2\Phi^*)=\bar D_{\dot\alpha}(4m\Phi+4\lambda\Phi^2)=0,}
$$
so applying the chiral constraint to the equation imposes no additional incompatible condition. This is the <holomorphic closure of chiral superfields>; a generic nonholomorphic right side would not satisfy it.
Both sides have zero <Grassmann parity> and are <Lorentz scalars>: the two dotted derivative indices are contracted with the invariant epsilon tensor, and $\Phi$ is a scalar <superfield>. Moreover, each covariant derivative anticommutes with every <supercharge>, so its squared contraction commutes with the <supercharges>. Hence the equation transforms covariantly under <supersymmetry>. It is the chiral field <equation of motion> of the <Wess–Zumino model> with <superpotential> $\mathcal W(\Phi)=m\Phi^2/2+\lambda\Phi^3/3$, for which $\mathcal W'=m\Phi+\lambda\Phi^2$, using the normalization in the PDF. With <mass dimensions> $[\Phi]=1$, $[\bar D]=1/2$, $[m]=1$ and $[\lambda]=0$, each side has dimension two. No component field <equations of motion> are needed for this consistency argument.
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