How to convert an expression from SOP to POS and back in Boolean Algebra?
I think the easiest way is to convert to a k-map, and then get the POS. In your example, you've got:

  \ xy
 z \  00    01    11    10
    +-----+-----+-----+-----+
 0  |     |  x  |  x  |  x  |
    +-----+-----+-----+-----+
 1  |     |     |     |  x  |
    +-----+-----+-----+-----+


In this case, excluding the left column gives (x+y), and excluding the two bottom middle boxes gives (z' + y'), giving an answer of (x + y)(z' + y')