= Solution
The <momentum space> <Feynman rules> use the <scalar propagator> and two <Dirac propagators>
$$
\boxed{\Delta_\phi(p)=\frac{i}{p^2-m^2+i0},\qquad
S_\chi(p)=\frac{i(\not p+m)}{p^2-m^2+i0},\qquad
S_\Psi(p)=\frac{i(\not p+M)}{p^2-M^2+i0}.}
$$
The cubic <scalar field> <interaction vertex> is $-i\lambda$. Each of the two flavour-changing <Yukawa coupling> <interaction vertices>, $\phi\bar\Psi\chi$ and $\phi\bar\chi\Psi$, has factor $-ig\,\delta_{ab}$ in <Dirac spinor> indices. A <fermion flow> arrow continues through each <interaction vertex>, but its species changes. There is no $\phi\bar\Psi\Psi$ or $\phi\bar\chi\chi$ <interaction vertex>.
Every <interaction vertex> carries $(2\pi)^4\delta^{(4)}(\sum p)$ with inward <four-momenta>. Integrate each independent <loop momentum> with $d^4\ell/(2\pi)^4$, divide by the <Feynman-diagram symmetry factor>, and include a minus sign and a <gamma-matrix trace> for each closed <fermion loop>. The $i0$ is the <Feynman i-epsilon prescription>.
For an amputated <scattering amplitude>, incoming <fermions> supply $u^s(p)$ and outgoing <fermions> supply $\bar u^s(p)$; incoming <antifermions> supply $\bar v^s(p)$ and outgoing <antifermions> supply $v^s(p)$. An external <real scalar field> particle supplies $1$. Use the standard <relativistic normalization of a one-particle state>; the external factors $1/\sqrt{2E}$ belong to that normalization and the <Lorentz-invariant phase-space measure>, rather than an additional factor in the invariant <scattering amplitude>.
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