access private member using template trick

You are not calling get from a! Actually what get return is a class pointer to a member inside A and type of it is int A::* so you need an instance of A to access that value.

For example let me play a little with your code:

struct A {
    A(int a):a(a) { }
    int b;
private:
    int a;
};
void test() {
    auto p = &A::b;
    std::cout << a.*p << std::endl;
}

Did I call p from inside a? a does not have p, this is exactly what happened in your code, get function return &A::a and you use a to read its value! that’s all, nothing is wrong and I think it will be compiled in all compilers.

One other question here is: Why C++ allow declaring template using private member of A. C++ standard say:

14.7.2p8 The usual access checking rules do not apply to names used to specify explicit instantiations. [Note: In particular, the template
arguments and names used in the function declarator (including
parameter types, return types and exception specifications) may be
private types or objects which would normally not be accessible and
the template may be a member template or member function which would
not normally be accessible.]

But if you try to instantiate or even typedef specified template then you get an error.
Let’s modify your example slightly:

struct A {
private:
    int a;
    friend void f();
};

// Explicit instantiation - OK, no access checks
template struct Rob<A_f, &A::a>;

// Try to use the type in some way - get an error.
struct Rob<A_f, &A::a> r;            // error
typedef struct Rob<A_f, &A::a> R;    // error
void g(struct Rob<A_f, &A::a>);      // error

// However, it's Ok inside a friend function.
void f() {
    Rob<A_f, &A::a> r;               // OK
    typedef Rob<A_f, &A::a> R;       // OK
}