= Solution
The last panel of the preceding process figure shows <deep inelastic scattering> through a virtual <photon>. The blob represents the electromagnetic <hadronic tensor> response, with all final hadronic states included. With charges factored into $e$ and the hadronic current as in the question, a suitable <scattering amplitude>, up to an irrelevant overall phase, is
$$
\boxed{\mathcal M_X=\frac{e^2}{Q^2}\overline u(p')\gamma_\mu u(p)
\langle X|J_h^\mu|H(P)\rangle,\qquad Q^2=-q^2.}
$$
The electron <fermion spin sum>, before averaging the two initial spin states, gives the following normalization of the <leptonic tensor>:
$$
\boxed{L_{\mu\nu}=\operatorname{Tr}[\not p'\gamma_\mu\not p\gamma_\nu]
=4(p'_\mu p_\nu+p'_\nu p_\mu-g_{\mu\nu}p\cdot p').}
$$
The initial electron spin average is a separate factor $1/2$. Equivalently one can absorb it into a tensor $L/2$, but the prefactor in the requested formula uses the unaveraged trace $L$. The massless Dirac equations imply $q^\mu L_{\mu\nu}=q^\nu L_{\mu\nu}=0$, the relevant <Ward identity>.
In the target rest frame the incident flux is $4p\cdot P=4ME$. This remains valid for a massive target because the electron is massless: the general flux $4\sqrt{(p\cdot P)^2-p^2P^2}$ reduces to that expression. The outgoing electron contributes $d^3p'/((2\pi)^3 2E')$, and the inclusive final-state sum is $4\pi W_H^{\mu\nu}$. Including the electron spin average therefore gives
$$
\begin{aligned}
\frac{d\sigma}{d^3p'}
&=\frac{1}{4ME}\frac{1}{(2\pi)^3 2E'}\frac{e^4}{(Q^2)^2}
\frac12L_{\mu\nu}(4\pi W_H^{\mu\nu})\\
&=\boxed{\frac{e^4}{(2\pi)^2\,8MEE'(Q^2)^2}
L_{\mu\nu}W_H^{\mu\nu}.}
\end{aligned}
$$
The implicit hadron spin average in $W_H$ must not be inserted a second time.
For an unpolarized target, <Lorentz invariance> permits tensor structures formed from $P$, $q$ and the metric. Hermiticity relates the two index orders. The electromagnetic response is parity even, excluding the pseudotensor $i\epsilon^{\mu\nu\rho\sigma}P_\rho q_\sigma$. A possible ordinary antisymmetric $P^\mu q^\nu-q^\mu P^\nu$ term is excluded by <current conservation>. The symmetric part initially has four structures, $g^{\mu\nu}$, $P^\mu P^\nu$, $q^\mu q^\nu$ and $P^\mu q^\nu+q^\mu P^\nu$. Imposing both current-conservation conditions leaves the <transverse decomposition of an unpolarized hadronic tensor>:
$$
W_H^{\mu\nu}=-\left(g^{\mu\nu}-\frac{q^\mu q^\nu}{q^2}\right)F_1
+\left(P^\mu-\frac{\nu}{q^2}q^\mu\right)
\left(P^\nu-\frac{\nu}{q^2}q^\nu\right)\frac{F_2}{\nu},
\qquad\nu=P\cdot q.
$$
For example, contracting $q$ into the first structure gives $(1+A)q^\mu$ in the printed parametrization, and contracting into either parenthesis gives $(1+B)\nu$ or $(1+C)\nu$. Since the two independent <deep-inelastic structure functions> are not generally linked by symmetry, transversality requires
$$
\boxed{A=B=C=-1.}
$$
There are only two invariant variables once $P^2=M^2$ is fixed; they may be chosen as $Q^2$ and <Bjorken x> $x=Q^2/(2\nu)$.
Now make the leading <massless collinear parton approximation>. In the hard parton calculation, use a lightlike collinear reference for $P$, neglecting target-mass effects of relative order $M^2/Q^2$ as well as quark masses. Literally combining a massive $P$ with an exactly massless constituent momentum $\xi P$ would be inconsistent. The exact rest-frame flux derived above does not require this approximation; it is the following partonic response that does.
Let $r=\xi P$ and $k=r+q$. Strip the quark charge out of the current, since its square is already the factor $Q_f^2$ in the parton convolution. Averaging the initial quark spin and summing the final spin gives the <spin-averaged electromagnetic parton tensor>
$$
S^{\mu\nu}=\frac12\operatorname{Tr}[\not r\gamma^\nu\not k\gamma^\mu]
=2(r^\mu k^\nu+r^\nu k^\mu-g^{\mu\nu}r\cdot k).
$$
The spatial momentum delta fixes $\boldsymbol k=\boldsymbol q+\xi\boldsymbol P$. The remaining one-particle <Lorentz-invariant phase space>, together with $1/(4\pi\xi)$ in the definition, supplies
$$
\frac{1}{4\pi\xi}\frac{(2\pi)^4}{(2\pi)^3\,2E_k}
\delta(E_q+\xi E_P-E_k)=\frac{\delta(E_q+\xi E_P-E_k)}{4\xi E_k}.
$$
Since $r^2=k^2=0$ and $r\cdot k=\xi P\cdot q$, the full parton response is
$$
\widetilde W^{\mu\nu}=\frac{\delta(E_q+\xi E_P-E_k)}{E_k}
\left[\xi P^\mu P^\nu+\frac{P^\mu q^\nu+q^\mu P^\nu}{2}
-\frac{\nu}{2}g^{\mu\nu}\right].
$$
Terms with a free $q$ index vanish when contracted with the conserved <leptonic tensor>. Thus \b[the part of $\widetilde W$ contributing to the cross-section is]
$$
\boxed{\widetilde W^{\mu\nu}\ \doteq\
\frac{\delta(E_q+\xi E_P-E_k)}{E_k}
\left[\xi P^\mu P^\nu-\frac{\nu}{2}g^{\mu\nu}\right],}
$$
where $\doteq$ denotes equality in that contraction, rather than an equality of the full tensors. The omitted terms are necessary for the full <hadronic tensor> to be transverse; the wording about contributing terms in the PDF is important here.
To prove the delta identity with the correct Jacobian, set $\omega=E_q+\xi E_P$. On physical support $E_k>0$ and $\omega>0$. Factoring $E_k^2-\omega^2=(E_k-\omega)(E_k+\omega)$ gives the <positive-energy on-shell delta function identity>
$$
\frac{\delta(\omega-E_k)}{E_k}=2\delta(E_k^2-\omega^2).
$$
This is restricted to the positive-energy branch; the other root must not be included. Using the spatial momentum constraint and $P^2=0$ in the leading hard approximation,
$$
E_k^2-\omega^2=-(q+\xi P)^2=Q^2-2\xi\nu=2\nu(x-\xi).
$$
The <Dirac delta function> change of variable therefore gives the <parton on-shell delta identity>
$$
\boxed{\frac{\delta(E_q+\xi E_P-E_k)}{E_k}
=2\delta(E_k^2-(E_q+\xi E_P)^2)
=\frac{\delta(x-\xi)}{\nu}.}
$$
Here $\nu>0$. This is the actual printed identity: the converted TeX's final $\delta(x)/E_k$ is a transcription error. If target mass is formally retained in the delta argument, it becomes $Q^2-2\xi\nu-\xi^2M^2$, so the stated simple root and Jacobian no longer apply.
Let $D(x)=\sum_fQ_f^2(q_f(x)+\overline q_f(x))$. The parton integral collapses at $\xi=x$, giving
$$
W_H^{\mu\nu}\doteq D(x)\left[\frac{x}{\nu}P^\mu P^\nu-\frac12g^{\mu\nu}\right].
$$
The <transverse decomposition of an unpolarized hadronic tensor>, in the same contraction, is $-g^{\mu\nu}F_1+P^\mu P^\nu F_2/\nu$. Comparing the two independent structures yields
$$
\boxed{F_1(x,Q^2)=\tfrac12D(x),\qquad F_2(x,Q^2)=xD(x),\qquad
F_2=2xF_1.}
$$
This derives the <Callan-Gross relation> from the spin-one-half trace and on-shell constraint. Antiquarks give the same symmetric response and therefore the sum $q_f+\overline q_f$. The result applies at leading order in the massless <parton model>; target masses, intrinsic transverse momentum and <Quantum chromodynamics> radiative effects can generate a nonzero <longitudinal deep-inelastic structure function> and modify the relation.
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