Role of diodes

Photodiodes can be used in two ways;
Either:
With an external reverse bias - in which case they conduct, passing a current proportional to the level of illumination,
or,
in "photovoltaic" mode, where they produce a voltage or current proportional to the illumination.
The voltage can never exceed the conduction voltage of the photodiode junction.

The circuit you show uses them in the second mode - as current sources.
With the opamp feedback holding the photodiode junction at 0V [it's a "virtual earth" in that opamp config], the diodes in the simulation become irrelevant, they can never conduct or have any function.

The photodiodes in a real circuit are the current sources in the sim.

Does capacitor means anything in the circuit?

In equivalent circuit of photodiode, ideal diode is one of the component. So what is the operation in that circuit.

In this equivalent circuit, How does ideal diode works?

I am trying to build the circuit as shown in the current subtracting design.
The Vout of opamp should be (I1 - I2) times Rf. But I am not getting the output as expected.
I am attaching my simulation and respective outputs.

As the diodes are not working like photodiodes and you have no current limiting, it will not do anything useful.

You should have got around 15mV out with the circuit before you added the extra voltage sources (in post #8).
That would be the same with or without the diodes and caps.