*a is pointer a, so what is **a ?

In the statement:

int *p; 

p is a pointer (not *p)
The variable declaration may be interpreted as follows:
*p is an int, hence p is a pointer to an int.

A similar reasoning gives

int ** q;

q is a pointer to a pointer to an int.

Generally speaking, the expressions *a, **a cannot be interpreted correctly without context.
For instance, the statement

printf("%d\n"), **q);

(if correct) tell us q is a double pointer.

Actually, *a is only syntactically correct if a itself is a pointer, and in that case *a can not be a simply because the types don't match, no matter what type a is defined to point to.

What you probably mean is the syntax used to define a (pointer) variable:

int f1() {
   int a; // this is a variable of type int
   a = 1;
   return a; // this will return the value 1
}
void f2() {
   int *a; // this is a pointer to a value of type int
   int b;
   a = &b; // now a points to the contents of b
   b = 2;
   return *a; // returns the value of b, which is 2
}
void f3() {
   int **a; // this is a pointer to a pointer to a value of type int
   int *b;
   int c = 3;
   b = &c; // ok, b points to the contents of c
   a = &b; // a now points to the contents of b which points to the contents of c
   return **a; // returns 3
}

It is called as double pointer,most of the times used to point to the dynamically created two dimensional arrays.
The following link has few more discussions:
http://www.daniweb.com/software-development/c/threads/69966[^]

**a is pointer of *a.
Otherway its a representation of 2 dimention pointer.
concider this:

a[] is array of a.
 AND
a[][] is array of a[].


suppose you have:
#define m=3
#define n=2
/////////////
...
int a[m][n];
int i,j;
for(i=0;i<m;i++)>
for(j=0;j<n;j++)>
{
///////
}

Like  this you can access:
int **a;
//use alloc or malloc to initialize the 2 dimentional pointer

for(i=0;i<m;i++)>
for(j=0;j<n;j++)>
{
///////
}