My pointer/addresses are wrong no matter what I try

now a,b and ptr are 3 different variables (regardless of their type)

So their address is different. You'll have the same values by not taking the address of pointer variables.

int a,*b;
int *ptr;

a = 123;
ptr = &a;
b = ptr;

printf("&a is %p\n",&a);
printf("ptr points to %p\n",ptr);
printf("b is %p\n",b);

I expected &b to show the same address as &a…

Having that expectation means, that you have a wrong mental model about how pointers work and what their semantics are in the C programming language. The key misconception seems to happen at these two lines of the original code fragment:

First you have

b = *ptr;

This line translates into *"copy the contents of the memory at address ptr into the variable b." The variable b never even gets into contact with the pointer itself. You could rewrite it perfectly fine to the same effect into

int tmp = *ptr;
b = tmp;

As a matter of fact, every modern C compiler out there will produce identical code for either case.

The second line where you have a misconception is this one

printf("&b is %p\n",&b);

specifically the effects of taking the address of the variable b, i.e. the result of &b. b is a completely independent variable, with its very own address. This address cannot be changed! The only thing that can be changed about a variable is its value.

Pointers are variables, too, and yes, a pointer does have an address itself. The value of a pointer is the address it points to. But like every variable, you cannot change the address where the variable is stored and when you assign an address to a pointer, you are changing its value. Hence when in your second code snipped you assign b = ptr; you're copying the value of ptr to b. After that, both ptr and b point to the same address, but these two pointers are two independent copies of the same value, where each copy is placed at a different place in memory. Naturally taking the address of b yields something different as the address of ptr.