Well, the photons are entangled in both the polarization states and the momentum, but this entanglement indeed doesn't do anything with respect to correlations between photon detectionprobabilities at a given place at Alice's and Bob's screens.
To see this note that the photons before the double slits are described by
$$|\Psi \rangle=\frac{1}{\sqrt{2}} [\hat{a}^{\dagger}(\vec{k},1) \hat{a}^{\dagger}(-\vec{k},-1) - \hat{a}^{\dagger}(\vec{k},-1) \hat{a}^{\dagger}(-\vec{k},1)] |\Omega \rangle.$$
For the double-slit experiment the polarizations and its entanglement doesn't play much of a role as long as you don't use it to imprint "which-way information" on the photons using the polarization (as e.g., used in Walborn's quantum-eraser experiment).
Let ##\vec{k}=k \vec{e}_z## and have put the Alice's screen to observe the photons with momentum ##+\vec{k}## behind her double slit at distance ##z=L_A## from these slits and the position on her screen ##x_A \vec{e}_x## an Bob's screen to observe the photons with momentum ##-\vec{k}## at ##z=-L_B## and the position on his screen at ##x_B \vec{e}_x## (with the slits of width ##b_A## and distance ##d_A## and ##b_B## and ##d_B## respectively infinite in the ##y## direction, so that you only get an interference pattern along the ##x## direction).
As explained on slide 3 of
https://itp.uni-frankfurt.de/~hees/publ/habil-coll-talk-en.pdf
The state after the slits is
$$|\Psi' \rangle=N_{\text{ESA}}(x_A)[1+\cos(\phi_{\text{A}}(x_A))N_{\text{ESB}}(x_B)[1+\cos(\phi_{\text{B}}(x_B)) |\Psi \rangle.$$
The only thing that happens is that you have a common position dependent factor which is due to the phase difference between the partial waves running through the one or the other slit of the double slits, leading to the "(1+cos)-factor" and the corresponding factor from the finite width of the single slits leading to the single-slit ##N_{\text{ES}}## factor. The interference pattern is then given by these amplitude factors squared, and they simply factorize for A's and B's photon. So there is no correlation for the detection probability at given places ##x_A## and ##x_B## for the single photons.
This simple calculation shows that the polarization state doesn't play any role here. It's only important that the single photons are coherent to get an interference pattern (in the here given idealized calculation with sharp plane-wave states, the photons are of course ideally coherent). The polarization state doesn't play any role in forming the detection probability and thus also the entanglement doesn't have any effect, i.e., there are no correlations between the detection positions of A's and B's entangled photons, as
@DrChinese says.
