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Martin is right. It is very simple to see it: just consider for a moment to add this extra superpotential term $$ \delta W(\Phi_i)=\frac{1}{2}m_{i j}\Phi_i \Phi_j $$ where $\Phi_i$ are chiral superfields. This extra $W$ corresponds to an extra lagrangian term $$ \delta\mathcal{L}=-\frac{1}{2}(\psi_i\psi_j+h.c)- \bar{\phi}_iM^2_{ij}\phi_j\qquad M^2_{ij}=\bar{m}_{ik}m_{jk}\,. $$ Therefore, starting with $W$ and adding a soft mass for the fermions $\delta\mathcal{L}_{\mathrm{soft}-\psi}=-\frac{1}{2}(\psi_i\psi_j+h.c)$ $$ W(\Phi)\rightarrow \mathcal{L}=\mathcal{L}_{SUSY}+\delta\mathcal{L}_{\mathrm{soft}-\psi} $$ is just the same as doing starting with $W+\delta W$ (with $\delta W$ in the first equation above) and add a soft mass terms for the scalars, $\delta\mathcal{L}_{\mathrm{soft}-\phi}=+\bar{\phi}_iM^2_{ij}\phi_j$ $$ W(\Phi)+\delta W(\Phi)\rightarrow \mathcal{L}=\mathcal{L}^\prime_{SUSY}+\delta\mathcal{L}_{\mathrm{soft}-\phi}=\mathcal{L}_{SUSY}+\delta\mathcal{L}_{\mathrm{soft}-\psi} $$