 Solution

Set $s=(p_1+p_2{)}^2=m_h^2$, so $p_1^2=p_2^2=0$ and $p_1\cdot p_2=s/2$. The physical polarization vectors obey ${ϵ}_1\cdot p_1={ϵ}_2\cdot p_2=0$. Write $X_{\mu \nu }\simeq Y_{\mu \nu }$ when they agree after contraction with these polarization vectors; terms proportional to $p_{1,\mu }$ or $p_{2,\nu }$ may then be discarded.

The gamma matrix trace identities kill the terms with an odd number of gamma matrices. The remaining three two-momentum terms and the term with three factors of $m$ areIn the $g_{\mu \nu }$ coefficient the terms linear in $l\cdot p_i$ cancel, leaving $m^2-l^2-s/2$. Among the other terms, the surviving mixed products of $l$ and $p_i$ also cancel. ConsequentlyIn the second Feynman diagram, exchange the external labels and change the dummy loop momentum $l\mapsto -l$. Its projected numerator and denominator become identical to those of the first Feynman diagram. HenceWith the common factor $8m$ extracted, the primitive integer coefficients areEquivalently, extracting only $m$ gives $(A,B,C)=(32,8,8)$ and prefactor $i{N}_c{\lambda }_t{e}^2{e}_t^2{m}_t/\sqrt{2}$. A common rescaling of all three coefficients changes the prefactor inversely. The physical scattering amplitude is ${ϵ}_1^{\mu }{ϵ}_2^{\nu }{\mathcal{M}}_{\mu \nu }^{\prime }$; the projected integrand is not an identity for an arbitrary uncontracted off-shell tensor.