Answer : 

      	111

      	222

      	333

      	344

      Explanation:

Let us consider the array and the two pointers with some address

   a	

0
1
2
3
4
         100      102      104      106      108

					     p

100
102
104
106
108
  			   1000    1002    1004    1006    1008

	     ptr	

1000
2000

After execution of the instruction ptr++ value in ptr
becomes 1002, if scaling factor for integer is 2 bytes. Now
ptr – p is value in ptr – starting location of array p,
(1002 – 1000) / (scaling factor) = 1,  *ptr – a = value at
address pointed by ptr – starting value of array a, 1002 has
a value 102  so the value is (102 – 100)/(scaling factor) =
1,  **ptr is the value stored in the location pointed by 
the pointer of ptr = value pointed by value pointed by 1002
= value pointed by 102 = 1. Hence the output of the firs
printf is  1, 1, 1.

After execution of *ptr++ increments value of the value in
ptr by scaling factor, so it becomes1004. Hence, the outputs
for the second printf are ptr – p = 2, *ptr – a = 2, **ptr = 2. 

After execution of *++ptr increments value of the value in
ptr by scaling factor, so it becomes1004. Hence, the outputs
for the third printf are ptr – p = 3, *ptr – a = 3, **ptr = 3. 

After execution of ++*ptr value in ptr remains the same, the
value pointed by the value is incremented by the scaling
factor. So the value in array p at location 1006 changes
from 106 10 108,. Hence, the outputs for the fourth printf
are ptr – p = 1006 – 1000 = 3, *ptr – a = 108 – 100 = 4,
**ptr = 4.