Following is another solution using k-map.
Given F$\left (x,y,z \right )$ = $\left ( x + y + z \right )\left (x’ + y\right )\left ( y’+z \right )$ in P.O.S form
Converting to S.O.P form we get
$\left (x’ y’ z’ \right )+ \left (x y’\right ) + \left ( yz' \right )$.
Now we can plot a k-map and put 0’s for the above S.O.P expression
To understand more about plotting max terms in a k-map, check https://www.allaboutcircuits.com/textbook/digital/chpt-8/minterm-maxterm-solution/
Now, we have marked the max terms in the k-map.
Now, objective is to find F’
From the above k-map, F’ can be easily computed by grouping 0s instead of 1s. Grouping 0s give the complement of F in S.O.P form. Now, check the options
Option #B is evident from the map by grouping 0’s ( we get S.O.P form )
Option #C is same as option #B but expressed in P.O.S form
Option #D is also possible if we don't do a grouping of 4 0’s
For Option #A , look at the k-map for 1s. Given F can be expressed as a sum of minterms = $\sum$( m1, m3, m7 )
For complement of F, find the missing minterms out of the 2^3=8 minterms possible.We have $\sum$( m0, m2, m4,m5, m6)
F’ = $\sum$(m0,m2,m4,m5,m6)
Expressing this in P.O.S form , $\prod$(M1,M3,M7) which is $\left (x +y’ + z’ \right )\left ( x+ y’ + z’ \right )\left ( x’ + y’ + z’ \right )$ which is not equivalent to option #A.
So,correct answers are options B, C, D