Differentiating a pointer from an array.

There is no way Fun may know that, unless the caller pass it the information.
The usual approach is (see, for instance GetWindowText[^] function): the caller pass the pointer to the buffer together with the size of the buffer itself.

With C++, is is quite easy to detect array on the stack using templates.... But type information must not be lost before being used. If you only want to support arrays, then you can delete the first function that works on pointers (or add a size parameter).

Alternatively, you can uses std::vector<int> instead (and have appropriate overload).

void Fun (int *p)
{
    // p is a pointer
}

template <int N> void Fun(int (&p)[N])
{
    // p is an array
}

int main()
{
    int a[5];
    Fun(a);    // Will call second function

    int *p = a;
    Fun(p);    // Will call first function
}

Generally if the code is complexe, then the template version should call a non-template version with an extra argument to avoid code-bloat and keep the convenience of not having to explicitly specify the size at the caller.

By the way first function should generally be deleted (or an extra size argument added) to avoid silent misuse.

Since the name of an array is converted into a pointer to the first element (ie. array == &(array[0])), there is a way, sort of, but it won't always work.
The sizeof operator returns the size of an array in bytes if fed an array name, or the size of a pointer if fed a pointer. So, provided that your array contains a different number of bytes to a pointer in your current system, you can tell that way.

Otherwise, I know of no reliable way to tell, in standard C or C++

An array variable name differs to a plain pointer in the following sense:

// array is not "int *" but "int * const", i.e. the address is const.
int a[] = {1, 2};
a++; // compiler error

Note: One might be tempted to overload F:
F(int *)
F(int * const) (identical to F(int[]))

Unfortunately, this int * const or int[] both translate to int * if it comes to function overlaod.

Cheers

Andi