First, let's get to the root of this problem, then I'll double back to your question. We are asked which of the two compounds is most acidic. As a side note there are various definitions of acidity, but note that broadly these theories describe acidity as linked to propensity to accept electron pairs in some form rather than general instability.
Considering neither II nor III have any immediately acidic functionalities (alcohol, carboxylic acid, sulfonic acid), we identify the alpha carbons of the carbonyl functionalities in each molecules as the most acidic hydrogens with pka ~20-25 (since all other hydrogens are alkyl (pKa ~50) and cannot form a stable conjugate base upon deprotonation).
With that in mind, the real difference between II and III is the identity of the carbonyl group. Comparing II to III we see II is an ester, with the carbonyl center bonded to an alkoxyl substituent and an alkyl substituent. Comparatively, III is a ketone, with its carbonyl center bound to to alkyl substituents. Recognizing that O lone pairs can hybridize to conjugate with the carbonyl center were C atoms cannot, we predict III to be more acidic.
The reasoning for this can be approached from a couple of angles, so here's two:
O lone pairs present in II destabilize the enol formed by deprotonation, making it more likely to donate electrons (I.E has more basic character than III), thus making its conjugate acid form less acidic (since weaker acids have conjugate bases that are stronger bases and vice versa)
O lone pairs can form an additional bond to the carbonyl carbon in a octet-satisfied resonance contributor, which greatly reduces the contribution of the analogous hyperconjugative resonance contributor with dissociation of H+. Note that this hyperconjugative contributor is a rather poor description in a thermodynamic sense, but it does play an effective stand in for a mechanism with a basic species.
Now to your actual question: You are correct with your analysis of the pi system of II, but I would note that your description is using atomic orbitals which can become very confusing in the world of ochem. In the hand-wavy hybridization terms, each oxygen would adopt an sp2 bonding orbital hybridization with unhybridized p orbitals participating in the pi-system. However, these pi-interactions can only occur if sequential atoms have unhybridized p orbitals.
This is due to the fact that the stability of pi systems arises from overlap of p orbitals. When two p orbitals are close enough to overlap, their wavefunctions can recombine into relatively lower energy (bonding) and relatively higher energy (antibonding) wavefunctions which become orbitals. This same recombination can occur for any number of p-orbitals, with the sole condition being that overlap actually occurs.
This can only have a fully stabilizing effect when atoms are ~1.20-1.5 angstroms apart as indicated by spectroscopic bond lengths. Overlap is feasible between single-bonded substituents (Like C-O in II), but impossible once the continuous system of unhybridized p orbitals is broken.
Thus while the O in II can directly conjugate its electrons with the carbonyl pi system by "rehybridizing" from the general expected sp3 hybridization to stabilize the system, in III there are no adjacent unhybridized p orbitals to interact with the carbonyl pi system. The only electronic interactions present between the substituent (cyclic ether) and the carbonyl system would be inductive/field effects.
While one could argue that a conformer of III could possibly involve delocalization of O p electrons into the carbonyl system, I would respond by noting that such a conformer
- Would have to have an unhybridized O p-orbital in order to be involved with pi interactions
- Would have to have a torsional angle Θ between the oxygen and the carbonyl group of Θ=/=0 in order for overlap to actually occur (Distance between orbitals r<~1.5).
However, this would mean that the oxygen's unhybridized p orbital and the carbonyl pi system would have different symmetry elements, meaning such a pi-interaction would be symmetry-forbidden
That last bit might be jumping the shark a little, but I hope this helps!