>i am needed to represent an array using pointers only
That suggests that you're required to use pointer arithmetic and dereferencing rather than subscripting. But, because the requirements clearly say "array", and not "dynamic array", or "simulated array" (granted, most teachers aren't intelligent enough to make the distinction), I would wager that you can still use an array and call it good.
To replace subscripting with pointer arithmetic, you need only realize that this:
array[index]
Is converted by the compiler internally to this:
*( array + index )
Now, let's shred your code. :)
>#include
This is a pointless include. You use getch, but only to "keep the window open" when running your program. That basically restricts the number of compilers you can run this code in to the ones that support getch in conio.h. Now, you could use getchar from stdio.h and have your code run everywhere. Using getch for this purpose is a naive rookie mistake that you would do well to rid yourself of as soon as possible.
>void main()
main returns int, there is no wiggle room here. If you want your code to be correct for all hosted implementations, main returns int. And when you say main returns int, you need to actually return a value too. 0 is a good choice for successful termination.
>b[0]=a[0];
Okay, and what does b point to again? If you said "I dunno", you get a cookie, because that's precisely where it points! Before using a pointer, you make sure it points to memory that you own, otherwise your program is probably undefined and seriously broken.
>printf("%s",a);
What makes you think that a contains a remotely reasonable string? a[0] is 2, a[1] is 2. Not only is that likely to have interesting output, you also failed to set a[3] to '\0' so that printf doesn't walk all through memory you don't own looking for a null character. Until you know how strings work, you would be better off initializing a string when you declare it:
#include <stdio.h>
int main ( void )
{
char a[] = "test";
char *b = a;
*( b + 2 ) = '!';
printf ( "%s\n", a );
printf ( "%s\n", b );
return 0;
}