210 CLA // clear accumulator
211 ADD 217 //add contents of location 217 i.e 1234 h to accumulator
212 INC // increment accumulator by 1 so accumulator contain 1235 h
213 STA 217 // store this value 1235 h at location 217
214 LDA 218 //load accumulator with content of 218 so accumulator now has 9CE2 h
215 CMA // complement (1’s complement )accumulator so accumulator now have 631D H
216 AND 217 // perform and operation with content of 217 i.e with 1235 h i.e and of 631D h & 1235 h which will be 0215 h
217 1234 H
218 9CE2 H
Hence option 4 is right ans