The rhs() routine expects an equation, and will not accept a set (even if its only member is an equation). So you could get your hands on that equation itself, or map the rhs() function over all the elements of the set.

> p1 := x^2+y;
                                        2
                                 p1 := x  + y

> e:=solve({p1}, {x});
                                    1/2             1/2
                      e := {x = (-y)   }, {x = -(-y)   }


> op(e[1]);
                                          1/2
                                  x = (-y)

> x1 := rhs(op(e[1]));
                                           1/2
                                 x1 := (-y)

> x1 := map(rhs,e[1]);
                                           1/2
                                x1 := {(-y)   }

> e[1][1];
                                          1/2
                                  x = (-y)

> x1 := rhs(e[1][1]);
                                           1/2
                                 x1 := (-y)

acer

Instead of fooling around with op, why not use the solution directly?  For example

x1 := eval(x, e[1]);

will work just as well. It will work over expressions in x too, which is the whole point of having the output of solve being in that format, it can be fed as input to eval as is.