struct point origin, *pp;

pp = &origin;
printf("origin is (%d,%d)\n", (*pp).x, (*pp).y);

The parentheses are necessary in (*pp).x because the precedence of the structure member operator . is higher then *. The expression *pp.x means *(pp.x), which is illegal here because x is not a pointer.
Pointers to structures are so frequently used that an alternative notation is provided as a shorthand. If p is a pointer to a structure, then

p->member-of-structure

refers to the particular member. So we could write instead
printf("origin is (%d,%d)\n", pp->x, pp->y);

Both . and -> associate from left to right, so if we have
struct rect r, *rp = &r;

then these four expressions are equivalent:
r.pt1.x
rp->pt1.x
(r.pt1).x
(rp->pt1).x