The 8051 (not the assembler) only has the instruction
movc a,@a+dptr. You are correct in assuming adding 0 to the dptr in this code does not affect the dptr - so why do it? Because that what the instruction does! We have no choice in the matter - there is no movc a,@dptr instruction.


We could have written your sample code like this:

mov dptr,#hello_str
call pstring

pstring:
mov r7,#0 ;reset index
ps1:
mov a,r7 ;get index into A
movc a,@a+dptr ;a= (dptr[R7])
jz ps_x ;if char == 0, exit
mov P1,a ;else copy to P1
inc r7 ;R7++
sjmp ps1 ;try for another character
ps_x:
ret

hello_str db "Hello world",0

In this instance the DPTR is the base pointer and R7 becomes the index. The limitation of this implementation is that a string can only be a maximum of 255 characters long and we use R7. My preferred way was your original code in using the DPTR as the index pointer.