Why is this way of calculating the diffraction pattern valid? I've seen that in some books (Fowles) the intensity of the diffraction pattern is calculated in the following way.

We place the source $S$ and the point at the screen $P$ in the line perpendicular to the aperture (second diagram). We calculate the intensity at $P$. 
Now, the rest of the pattern is obtained by displacing the aperture keeping $S$ and $P$ fixed (this would mean that we change the integration limits).
It is supposed that this method should be equivalent to keeping $S$ and the aperture fixed and moving $P$ (first diagram), which is what we really want to calculate.
But how can they both be equivalent? Are we making any approximations?
I've read in my note that this is done to simplify calculations, but no justification is given.
 A: In Fresnel diffraction, you are evaluating the contribution of every possible ray from source to screen by computing the relative phase shift for each ray. The method you show is only valid if the distance from source S to aperture Q is much larger than the distance from Q to P: that is the only condition in which small lateral displacements of S relative to Q don't affect the result (it will result in a small phase gradient across the aperture; that has to be negligible compared to the shift due to P).
So I disagree with @Nordik's answer: I don't think it's the distance between S and P that matters; it's the difference in distance such that SQ >> QP.
A: The difference is negligible as long as plane displacements are much less than distance S to P.
As you've mentionned, we are talking about Fresnel diffraction here.
And one of its requirements is the following:
$\rho\ll z,$
$\rho$ is a difference in plane coordinates between points S and P  $(\rho=\sqrt{(x_S-x_P)^2+(y_S-y_P)^2})$,
$z$ is a distance between S plane and P plane $(z=z_S-z_P)$.
Strictly speaking two approaches will give you different results. But the difference will be negligible as long as either aperture displacement or P displacement is of magnitude of $\rho$.
