The checking for NULL trick only works for NULL terminated strings. For a numeric array you'll have to pass in the size too.

void printArray(int *ptr, size_t length);            
{         
    //for statement to print values using array             
    size_t i = 0;
    for( ; i < length; ++i )      
    printf("%d", ptr[i]);        
}   

 void printString(const char *ptr);            
{         
    //for statement to print values using array             
    for( ; *ptr!='\0'; ++ptr)        
    printf("%c", *ptr);        
}         

int main()    
{    
    int array[6] = {2,4,6,8,10};     
    const char* str = "Hello World!";
    printArray(array, 6);    
    printString(str);
    return 0;     
}

fixing your many errors and re-editing to use strlen:

#define LENGTH 8
#include <stdio.h>
#include <string.h>
int main()
{
    typedef char Arr[LENGTH];
    Arr test  = {'1','0','1'};
    Arr* a = &test;
    size_t len = strlen(test);
    for(int i=0;i<len;i++){
        printf("%c ", (*a)[i]);
    }
}

Let's start from this line

int(*p)[10] = arr;

The array arr used as an initializer is implicitly converted to a pointer to its first element of the type int *.

However the initialized pointer p has the type int ( * )[10] and there is no implicit conversion between these two pointers.

So the compiler should issue at least a warning.

You have to write

int(*p)[10] = &arr;

On the other hand, values of the expressions arr and &arr are equal each other. It is the initial address of the extent of memory occupied by the array. (The address of the first element of an array is equal to the address of the array as a whole object).

So the function accepts a pointer of the type int ( * )[10] that points to an array.

Dereferencing the pointer in the expression (*p) you get an object of the type int[10] (the array pointed to by the pointer) that is an object of the array type.

So you can apply the subscript operator (*p)[i] to access elements of the array.