The status register will be pulled with the break flag and bit 5 ignored.
As far PLA instruction is concerned, it is clear that status bit N and Z will be affected by the operation, but I am unable to understand which bits will be affected by the PLP instruction.
The status register only really has 6 bits - C, D, I, N, V, Z - and they are all affected by pulling it from the stack, each being set to the value of one of the bits in the value pulled. The values of bits 4 and 5 in the byte from the stack are discarded.
The status register will be pulled with the break flag and bit 5 ignored.
As far PLA instruction is concerned, it is clear that status bit N and Z will be affected by the operation, but I am unable to understand which bits will be affected by the PLP instruction.
As the others noted, bits 4 and 5 are not “real” bits in the 6502/65C02—and in the 65C816 when the latter is operated in emulation mode. Bit 4 only exists in the stack copy of the status register (SR) that is pushed during a response to an IRQ or as part of executing a BRK (software interrupt) instruction. If an IRQ is being processed, bit 4 will be clear, otherwise bit 4 will be set. This behavior enables the IRQ/BRK service routine to distinguish between the two interrupt types.
When SR is pulled from the stack with a PLP instruction, bits 4 and 5 will not be affected by whatever is on the stack. The sequence PHP - PLA will result in bits 4 and 5 always being set in the accumulator copy of SR.
For detailed information on the 65xx instruction set, get yourself a copy of “Programming the 65816, Including the 6502, 65C02, and 65802” by Eyes & Lichty. I usually caution people about getting this kind of information from arbitrary websites, as such sites may propagate errors posted on other sites. The Eyes & Lichty manual may be considered authoritative for all permutations of the 6502.