How to determine no of bytes of an instruction of 8085 microprocessor?
The instruction format depends on the addressing mode.

Register-to-register instructions only need to know the opcode, the source register, and the destination register. Usually that much information can be encoded in one byte.

A Move-Immediate instruction needs 8 bits to store the immediate 8-bit data, plus the opcode and destination register, so that is a two-byte instruction.

A 16-bit move-immediate instruction (e.g. loading the HL register pair) requires 16-bits, plus the opcode and destination register encoding, so that is a three-byte instruction.