= Solution
A free contraction must connect one $\phi$ and one $\phi^*$ <field>. In <momentum> space,
$$
\boxed{\langle T\phi(p)\phi^*(-p)\rangle_0=\frac{i}{p^2-m_0^2+i0},\qquad
\langle T\phi\phi\rangle_0=\langle T\phi^*\phi^*\rangle_0=0.}
$$
Represent the nonzero <propagator> by a directed line so the two conjugate ends remain distinguishable; an undirected line alone would incorrectly allow same-charge contractions. Choose one consistent arrow convention, for example pointing from the starred end to the unstarred end.
The interaction contains two identical $\phi$ <fields> and two identical $\phi^*$ <fields>. Their $2!2!$ assignments cancel the coefficient $1/4$, giving
$$
\boxed{v_{\phi\phi\phi^*\phi^*}=-i\lambda_0.}
$$
The vertex has two incoming and two outgoing arrows. Include $(2\pi)^d\delta^d(\sum p_i)$ at every vertex, integrate $d^dk/(2\pi)^d$ for each independent loop, and divide by the oriented graph's <symmetry factor>. There is no fermion-loop sign.
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