I am a bit confused by how Peskin & Schroeder describe the corrections to the two point function of the linear sigma model from the second functional derivative of the effective action. Without going into too much details on p386 (Eq. 11.103) they derive the second functional derivative of the functional determinant and find \begin{align} \frac{i}{2} \frac{\delta ^2}{\delta \phi^l (w )\delta \phi^k (z)} &\log \det [-i \mathcal{D}] =-\lambda ( \delta^{kl}\delta^{ij} + \delta^{jl}\delta^{ik} + \delta^{il}\delta^{jk}) \delta (z-w) \left( \mathcal{D}^{-1}\right) ^{ij} (z,z)\nonumber\\ & +2i \lambda^2 \left[ \phi^k(z) \delta^{ij} + \phi^j (z) \delta^{ik}+ \phi^i(z) \delta^{jk}\right] \nonumber\\ &\times \left( \mathcal{D}^{-1}\right) ^{im}(z,w) \left[ \phi^l(w) \delta^{mn} + \phi^m (w) \delta^{ln}+ \phi^n(w) \delta^{lm}\right] \left( \mathcal{D}^{-1}\right) ^{nj}(w,z). \tag{11.103} \end{align} I have dropped the subscript $_\text{cl}$ for convenience. So far so good. They then say that it is related to the inverse propagator and that we recognise the one-loop diagrams. I have added these diagrams at the end of the question (apologies for the formatting, I am not sure how to do it better).
I really don't understand how these are related. The first line of the equation has no field, so I expect no external line; I would have expected a vacuum bubble. The second and third line have a $\lambda^2$ factor so I expect two vertices, but the first diagram only has one vertex. They also have two $\phi$ fields and two propagators, so again how does the first diagram fit in? Or does the first diagram correspond to the first line of the equation, but then where do the external lines come from?
I am probably missing something simple here, but any help to put me on the right track would be much appreciated.