= Solution
Let the source remain at $(0,0,-p)$ and let the observation point be $(X,Y,q)$. The <paraxial approximation> to the source-to-aperture-to-observer <optical path length> is
$$
p+q+\frac{x^2+y^2}{2p}+\frac{(X-x)^2+(Y-y)^2}{2q}.
$$
Completing the squares gives
$$
p+q+\frac{X^2+Y^2}{2(p+q)}
+\frac{(x-x_0)^2+(y-y_0)^2}{2R_{\rm eff}},\qquad
x_0=\frac{pX}{p+q},\quad y_0=\frac{pY}{p+q}.
$$
The point $(x_0,y_0)$ is where the straight source-observer line intersects the aperture. Define the four dimensionless edge positions
$$
u_{x,\pm}=\sqrt{\frac{2}{\lambda R_{\rm eff}}}\left(\pm\frac d2-x_0\right),\qquad
u_{y,\pm}=\sqrt{\frac{2}{\lambda R_{\rm eff}}}\left(\pm\frac d2-y_0\right).
$$
On the <Cornu spiral> $F(u)=C(u)+iS(u)$, draw the directed chord from $F(u_{x,-})$ to $F(u_{x,+})$, and a second directed chord from $F(u_{y,-})$ to $F(u_{y,+})$. Denote their complex differences by $D_x$ and $D_y$. The <complex amplitude> is proportional to their product: multiply their lengths and add their arguments, including the common propagation <phase factor>.
More explicitly, normalize by the unobscured field $\psi_0(X,Y)$ at this same observation point. The constant <wave phase> from completing the square is then already included in $\psi_0$, and
$$
\boxed{\frac{\psi(X,Y)}{\psi_0(X,Y)}=\frac{D_xD_y}{2i},\qquad
\frac{I(X,Y)}{I_0(X,Y)}=\frac{|D_x|^2|D_y|^2}{4}.}
$$
Each directed chord is a one-dimensional <Fresnel diffraction> factor. One must multiply the two factors, rather than add the chords or use just one chord length for the square aperture. Setting $X=Y=0$ recovers $D_x=D_y=2F(w)$ and the on-axis answer.
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