We know
logical address contains: page number and page offset (page size)
physical address contains : number of frames and frame size
Logical address space contains 32 pages and to represent 32 pages we need 5 bits $2^5=32$.
Each page contains 2048 words therefore the offset or "d" parameter can be represented by 11 bits $2^{11}=2048$.
So total number of bits to represent logical address : $5+11=16$ bits.
Similarly 64 frames can be represented by 6 bits $2^6=64$ and offset will be same (because page size is equal to frame size) i.e. 11 bits.
The total number of bits to represent physical address: $6+11=17$ bits.