= Solution
Write $p$ and $q$ for the axial source and observation distances, and put $R_{\rm eff}=pq/(p+q)$. The aperture is $-d/2<x,y<d/2$. Under the <Fresnel diffraction>, the spherical incident <wave amplitude> is effectively constant over this small aperture, while its quadratic <wave phase> must be retained. Therefore the two transverse <integrals> separate:
$$
\psi_P=K\left[\int_{-d/2}^{d/2}\exp\left(\frac{i\pi x^2}{\lambda R_{\rm eff}}\right)dx\right]^2,
$$
where $K$ contains the source strength, the propagation distances and a common <phase factor>.
Use the dimensionless variable $u=x\sqrt{2/(\lambda R_{\rm eff})}$ and define the <optical normalization of Fresnel integrals>
$$
F(w)=C(w)+iS(w)=\int_0^w e^{i\pi u^2/2}\,du,\qquad
w=\frac d2\sqrt{\frac{2}{\lambda R_{\rm eff}}}.
$$
The oddness of $F$ makes the bracket $\sqrt{2\lambda R_{\rm eff}}F(w)$, so
$$
\boxed{\psi_P=2K\lambda R_{\rm eff}[C(w)+iS(w)]^2.}
$$
This <square-aperture Fresnel diffraction> normalization is especially useful for comparing with no aperture. In the same <Fresnel diffraction>, the unrestricted one-dimensional <Fresnel integral> is $\sqrt{\lambda R_{\rm eff}}e^{i\pi/4}$; its square is $i\lambda R_{\rm eff}$. Thus, writing $\psi_0$ for the unobscured field at the same point,
$$
\boxed{\frac{\psi_P}{\psi_0}=-2i[C(w)+iS(w)]^2,\qquad
\frac{I_P}{I_0}=4[C(w)^2+S(w)^2]^2.}
$$
The normalization has the correct no-aperture limit because $C(\infty)=S(\infty)=1/2$ in this convention.
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