Would there be fringe pattern in this arrangement? Figure shows a standard two slit
arrangement with slits S1, S2. P1, P2 are the two
minima points on either side of P.
At P2 on the screen, there is a hole and behind P2
is a second 2- slit arrangement with slits S3, S4
and a second screen behind them. 

Would there be a fringe pattern on the second screen?
I think no because the only way light can get to the slits S3 and S4 is through the hole at P2. But, its a minima there so no light passes through S3 and S4 and thus no fringe pattern on the second screen.
Is my thinking correct?
 A: In interference and diffraction, light energy is redistributed. If it reduces in one region, producing a dark fringe, it increases in another region, producing a bright fringe. There is no gain or loss of energy, which is consistent with the principle of conservation of energy.
If you consider a point where there is destructive interference, there is a dark fringe. So, "no more light energy" is passing through it, provided amplitude of light waves undergoing destructive interference is same or width of both the slits is same. Thus, we can't expect fringe pattern in the second screen.    
If amplitudes of light waves undergoing destructive interference at the point is not same, or if width of two slits is not same, intensity of light at the point of destructive interference will not be zero, then we can expect fringe pattern in the second screen.
Good comment by Carl Witthoft made me to stop answering this post for a while. In talbot effect there is only one diffraction grating. Later at regular distances from the grating, the light diffracted through it forms a nearly perfect image of the grating itself. But there are diffraction grating in between. It is totally a different wonderful concept. If interested I would suggest one to read this article: Rolling out the (optical) carpet: the Talbot effect.
A: The answer, as others have siad, is no.  You will not get an interference pattern at the second screen if the single slit is placed at a dark fringe and is sufficiently narrow.  You would get an interference pattern at the second screen if the single slit were placed at a bright fringe though.  Let me explain why.
First of all, it is important to keep in mind that interference happens wherever you place your detector and not before.  So you are right to question whether or not there will be an interference pattern at the second screen, the answer wasn't immediately obvious to me either.  
What it means to say that the interference doesn't happen until you've reached the detector is this: we have to trace both beams through the entire apparatus before asking how they interfere.  What I will try to show is that the phase difference between the two light sources, $S_1$ and $S_2$, is encoded at the single slit, $P_2$, since they traverse identical path lengths afterwards; both rays passing through both of the second pair of slits $S_3$ and $S_4$.
According to Huygen's principle each slit becomes a source of spherical waves with the phase of the incident wave encoded in them.  Lets think about the case where $S_1$ is blocked.  Slit $S_2$ will produce spherical waves which make it to $P_2$.  $P_2$ will produce spherical waves which make it to $S_3$ and $S_4$, and the screen will show a typical double slit interference pattern produced by slits $S_3$ and $S_4$.  The same thing will occur if we block $S_2$ and follow the spherical waves from $S_1$.
But what happens if both $S_1$ and $S_2$ are unblocked?  Since the light from both slits traverses the same path after $P_2$, with both going through $S_3$ and $S_4$, the only phase difference between the two is the phase picked up before getting to $P_2$.  But we chose this phase to be $\pi$ by placing $P_2$ at a dark fringe.  So the only phase difference at the final screen will be the phase difference picked up by the first screen.
You can think of it as if the two slits would individually produce their own double slit pattern, but the two double slit patterns interfere destructively!
