Char size = 1 Byte
Word size = 36 bits
Double word size = 72 bits = 9 Bytes.
Characters are packed into double words so that no space is lost (because 36 bits is Not a multiple of 8 But 72 is a multiple of 8)
If the string does not occupy an integral number of double words, enough binary zeros are appended to reach the next single word boundary. This means that memory is word aligned i.e. Next string starts at the beginning of a new word.
For example, if K=1 i.e. One character string S then we need to start next string from next word onwards and for this string S, we need to append zeroes to make it reach the next single word boundary. So, when K=1 then we need 1 word.
So, for K=1, answer will be 1. Hence we can eliminate option A,B,C. So, answer is option D.
Option C would be true if the memory was "double word aligned". In that case, for K=1, we would need 2 words.