= Solution
Let $\mathcal V_4=i\Gamma^{(4)}$ denote the amputated Feynman vertex, with tree term $v_4=-i\lambda$. A one-loop <one-particle-irreducible Feynman diagram> has $V_3+2V_4=4$, allowing three families: two quartic vertices, two cubic vertices and one quartic vertex, or four cubic vertices. Define
$$
I_4(P,Q,R,S)=\int_kD(k)D(k+P)D(k+P+Q)D(k-S),\quad P+Q+R+S=0.
$$
Using $I_2,I_3$ from part c and writing $P_{ij}=p_i+p_j$, the complete one-loop result is
$$
\boxed{\begin{aligned}
\mathcal V_{4,\,1\text{-loop}}={}&\frac{v_4^2}{2}
\left[I_2(P_{12})+I_2(P_{13})+I_2(P_{14})\right]\\
&+v_3^2v_4\sum_{1\leq i<j\leq4}I_3(P_{ij},p_k,p_l)\\
&+v_3^4\left[I_4(p_1,p_2,p_3,p_4)+I_4(p_1,p_2,p_4,p_3)+I_4(p_1,p_3,p_2,p_4)\right].
\end{aligned}}
$$
In each term of the middle sum, $k,l$ are the two complementary indices. Their order does not change the integral after a loop-<momentum> redefinition.
The first line contains three bubble channels, one for each partition of the four labelled external legs into two pairs. Each has $S=2$ from interchanging its two identical internal lines. In the second line, the quartic vertex carries a chosen pair of external legs, giving $\binom42=6$ mixed triangles. Each has $S=1$. In the last line, cyclic orderings of four labelled external legs are identified under rotation and reversal, leaving $(4-1)!/2=3$ boxes, again with $S=1$ each. The figure in part b shows one representative of each family and its multiplicity. External-leg corrections and tadpoles connected by bridges are excluded because $\Gamma^{(4)}$ is a <one-particle-irreducible vertex>. All integrals are left unevaluated, as appropriate here.
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