Doing "man malloc" gives all the required information. malloc just allocates the required amount of memory as mentioned to its argument. The typecast (int*) just does explicit typecasting - this does not change any size and type of p is "int *" as that is how it is defined and size of "int *" or any pointer is as defined on the system - 8 on 64 bit systems.
SYNOPSIS
#include <stdlib.h>
void *malloc(size_t size);
void free(void *ptr);
void *calloc(size_t nmemb, size_t size);
void *realloc(void *ptr, size_t size);
DESCRIPTION
The malloc() function allocates size bytes and returns a pointer to the allocated memory. The memory is not initialized. If size is 0, then malloc() returns either NULL,
or a unique pointer value that can later be successfully passed to free().
The free() function frees the memory space pointed to by ptr, which must have been returned by a previous call to malloc(), calloc(), or realloc(). Otherwise, or if
free(ptr) has already been called before, undefined behavior occurs. If ptr is NULL, no operation is performed.
The calloc() function allocates memory for an array of nmemb elements of size bytes each and returns a pointer to the allocated memory. The memory is set to zero. If nmemb
or size is 0, then calloc() returns either NULL, or a unique pointer value that can later be successfully passed to free().
The realloc() function changes the size of the memory block pointed to by ptr to size bytes. The contents will be unchanged in the range from the start of the region up to
the minimum of the old and new sizes. If the new size is larger than the old size, the added memory will not be initialized. If ptr is NULL, then the call is equivalent to
malloc(size), for all values of size; if size is equal to zero, and ptr is not NULL, then the call is equivalent to free(ptr). Unless ptr is NULL, it must have been
returned by an earlier call to malloc(), calloc() or realloc(). If the area pointed to was moved, a free(ptr) is done.