Kmap

w'x' 
w'x 
wx 
wx' 
y'z' 
1 
1 

1 
y'z 

1 
1 
1 
yz 

1 
1 

yz' 




So, minimized expression will be
$xz + w'y'z' + wx'y'$ which is Q. From the Kmap, we can also get P and R. So, only S is NOT equivalent to $f$.
http://www.eecs.berkeley.edu/~newton/Classes/CS150sp98/lectures/week4_2/sld011.htm
Go with Minterm representation of each option, (Note that order of W,X,Y,Z is should be preserved. )
P : X' Y' Z' + W' X Y' + W Y' Z + X Z
= () X' Y' Z' + W' X Y' () + W () Y' Z + () X () Z
= () 000 + 010() + 1 () 01 + () 1 () 1
= 0000 + 1000 + 0100 + 0101 + 1001 + 1101 + 0101+0111+1101+1111
= 0 + 8 + 4 + 5 + 9+13 + 5+7+13+15
= ∑ m(0,4,5,7,8,9,13,15)
Q : W' Y' Z' + W X' Y' + XZ
= W' () Y' Z' + W X' Y' () + () X () Z
= 0 () 0 0 + 100 () + () 1 () 1
= 0000+0100 + 1000+1001 + 0101 + 0111+1101+1111
= 0+4 + 8+9 + 5+7+13+15
= ∑ m(0,4,5,7,8,9,13,15)
R : W' Y' Z' + W X' Y' + XYZ + X Y' Z
= W' () Y' Z' + W X' Y' () + () X Y Z + () X Y' Z
= 0 () 0 0 + 100 () + () 111 + () 101
= 0000+ 0100 + 1000+1001 + 0111+1111 + 0101 + 1101
= 0+4 + 8+9 + 7+15 + 5+13
= ∑ m(0,4,5,7,8,9,13,15)
S : X' Y' Z' + W X' Y' + W' Y
= () X' Y' Z' + W X' Y' () + W' () Y ()
= () 000 + 100 () + 0 () 1 ()
= 0000+1000 + 1000+1001 + 0010+0011+0110+0111
= 0+8 + 8+9 + 2+3+6+7
= ∑ m(0,2,3,6,7,8,9)