What does getting the address of an array variable mean?
First a little reminder (or something new if you didn't know this before): For any array or pointer p and index i the expression p[i] is exactly the same as *(p + i).
Now to hopefully help you understand what's going on...
The array a in your program is stored somewhere in memory, exactly where doesn't really matter. To get the location of where a is stored, i.e. get a pointer to a, you use the address-of operator & like &a. The important thing to learn here is that a pointer by itself doesn't mean anything special, the important thing is the base type of the pointer. The type of a is int[4], i.e. a is an array of four int elements. The type of the expression &a is a pointer to an array of four int, or int (*)[4]. The parentheses are important, because the type int *[4] is an array of four pointers to int, which is quite a different thing.
Now to get back to the initial point, that p[i] is the same as *(p + i). Instead of p we have &a, so our expression *(&a + 1) is the same as (&a)[1].
Now that explains what *(&a + 1) means and what it does. Now let us think for a while about the memory layout of the array a. In memory it looks something like
+---+---+---+---+ | 0 | 1 | 2 | 3 | +---+---+---+---+ ^ | &a
The expression (&a)[1] treats &a as it was an array of arrays, which it definitely isn't, and accessing the second element in this array, which will be out of bounds. This of course technically is undefined behavior. Let us run with it for a moment though, and consider how that would look like in memory:
+---+---+---+---+---+---+---+---+ | 0 | 1 | 2 | 3 | . | . | . | . | +---+---+---+---+---+---+---+---+ ^ ^ | | (&a)[0] (&a)[1]
Now remember that the type of a (which is the same as (&a)[0] and therefore means that (&a)[1] must also be this type) is array of four int. Since arrays naturally decays to pointers to its first element, the expression (&a)[1] is the same as &(&a)[1][0], and its type is pointer to int. So when we use (&a)[1] in an expression what the compiler gives us is a pointer to the first element in the second (non-existing) array of &a. And once again we come to the p[i] equals *(p + i) equation: (&a)[1] is a pointer to int, it's p in the *(p + i) expression, so the full expression is *((&a)[1] - 1), and looking at the memory layout above subtracting one int from the pointer given by (&a)[1] gives us the element before (&a)[1] which is the last element in (&a)[0], i.e. it gives us (&a)[0][3] which is the same as a[3].
So the expression *(*(&a + 1) - 1) is the same as a[3].
It's long-winded, and passes through dangerous territory (what with the out-of-bounds indexing), but due to the power of pointer arithmetic it all works out in the end. I don't recommend you ever write code like this though, it needs people to be really know how these transformations work to be able to decipher it.
&a + 1 will point to the memory immediately after last a element or better to say after a array, since &a has type of int (*)[4] (pointer to array of four int's). Construction of such pointer is allowed by standard, but not dereferencing. As result you can use it for subsequent arithmetics.
So, result of *(&a + 1) is undefined. But nevertheless *(*(&a + 1) - 1) is something more interesting. Effectively it is evaluated to the last element in a, For detailed explanation see https://stackoverflow.com/a/38202469/2878070. And just a remark - this hack may be replaced with more readable and obvious construction: a[sizeof a / sizeof a[0] - 1] (of course it should be applied only to arrays, not to pointers).
Let's dissect it.
a has type int [4] (array of 4 int). It's size is 4 * sizeof(int).
&a has type int (*)[4] (pointer to array of 4 int).
(&a + 1) also has type int (*)[4]. It points to an array of 4 int that starts 1 * sizeof(a) bytes (or 4 * sizeof(int) bytes) after the start of a.
*(&a + 1) is of type int [4] (an array of 4 int). It's storage starts 1 * sizeof(a) bytes (or 4 * sizeof(int) bytes after the start of a.
*(&a + 1) - 1 is of type int * (pointer to int) because the array *(&a + 1) decays to a pointer to its first element in this expression. It will point to an int that starts 1 * sizeof(int) bytes before the start of *(&a + 1). This is the same pointer value as &a[3].
*(*(&a + 1) - 1) is of type int. Because *(&a + 1) - 1 is the same pointer value as &a[3], *(*(&a + 1) - 1) is equivalent to a[3], which has been initialized to 3, so that is the number printed by the printf.