= Solution
Assume the standard canonical kinetic action $\int d^4\theta\,\Phi^\dagger\Phi$. A constant <field redefinition> $\Phi=\Psi+s$ is invertible and preserves <chirality>. Its kinetic action differs by $s\Psi^\dagger+s^*\Psi+|s|^2$, whose full <superspace> integrals vanish up to a boundary term: a chiral or antichiral field has no independent full-superspace component. Equivalently, spacetime derivatives of a constant are zero. There is no nontrivial Jacobian for this translation of the field. Thus physical predictions are unchanged when all parameters and the coordinate of the vacuum are transformed together.
The transformed <superpotential> is
$$
W(\Psi+s)=A+\left(\kappa+ms+\frac g2s^2\right)\Psi
+\frac{m+gs}{2}\Psi^2+\frac g6\Psi^3,
$$
where $A=W(s)$. Removing the linear term means choosing a critical point $W'(s)=0$, not merely choosing $s=-\kappa/m$ while ignoring the cubic term. For $g\ne0$,
$$
\boxed{s=\frac{-m\pm\sqrt\Delta}{g},\qquad
\Delta=m^2-2g\kappa,\qquad M:=m+gs=\pm\sqrt\Delta.}
$$
The resulting <superpotential> is $A+M\Psi^2/2+g\Psi^3/6$. Its constant term has no effect in global <supersymmetry>, since its chiral <superspace> integral is zero. It would matter in <supergravity>, which is not the model here.
A real shift is physically allowed, but it does not always suffice to remove the linear term. With real couplings it requires $\Delta\geq0$ for $g\ne0$. For example $m=0$, $\kappa=1$, $g=2$ gives $W'(s)=1+s^2$, which has no real root. A complex shift $s=\pm i$ is still a valid redefinition of the complex chiral scalar and removes the term. Thus the printed instruction needs either the real-root hypothesis or permission for a complex shift.
If $g=0$ and $m\ne0$, choose $s=-\kappa/m$. If $g=m=0$, no shift removes a nonzero $\kappa$; if $\kappa=0$ it is already absent. These exceptional branches will also distinguish the supersymmetry-breaking possibilities below.
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