The question is indirctly asking for static hazard in the circuit  that is output becoming 1 momentarily when it is supposed to be 0 or vice verse.
Here $f(w,x,y,z) = \sum (5,7,11,12,13,15)$
So, Kmap will be

y'z' 
y'z 
yz 
yz' 
w'x' 




w'x 

5 
7 

wx 
12 
13 
15 

wx' 


11 

So, its minimized sum of products expression will be $xz + wxy + w'yz$. Since all the minterms are overlapping, there is no chance of static hazard here.
Now, lets consider the options one by one:
A. $\overline{w}xz,wx\overline{y},x\overline{y}z,xyz,wyz$

y'z' 
y'z 
yz 
yz' 
w'x' 




w'x 

5 
7 

wx 
12 
13 
15 

wx' 


11 

Chance of static hazard
Here, when $y$ changes from $0$ to $1$, the GATE for $wyz$ should give 1 (from earlier 0, assuming $w=z=1$) and that of $wy'z$ should give 0 (from earlier 1). But there is a possibility of circuit giving 0 (static 1 hazard) momentarilly due to gate delays. In order to avoid this, we must add a gate with $wxz$ also which ensures all adjacent blocks in Kmap are overlapped.
We can also have the sum of products as $xy'z + w'yz + wxy$ which is of course not minimal. But this also has hazard as when $w$ changes from $0$ to $1$, the GATE for $wxy$ should give 1 (from earlier 0) and that of $w'yz$ should give 0 (from earlier 1). But there is a possibility of circuit giving 0 momentarilly (assuming $x=0$). This hazard can be avoided by adding $xyz$ and similarly we need $w'xz$ and $wxz$. Thus we get $ xy'z + w'yz + wxy + xyz + w'xz + wxz$
Now, each of these term will correspond to a GATE when the circuit is implemented. What we require is to avoid the output to change when any of the input literal changes its value (hazard).
B. $\overline{w}xz,wx\overline{y},x\overline{y}z,xyz,wyz$

y'z' 
y'z 
yz 
yz' 
w'x' 




w'x 

5 
7 

wx 
12 
13 
15 

wx' 


11 

Chance of static hazard
Here, when $y$ changes from $0$ to $1$, the GATE for $wyz$ should give 1 (from earlier 0, assuming $w=z=1$) and that of $wy'z$ should give 0 (from earlier 1). But there is a possibility of circuit giving 0 (static 1 hazard) momentarilly due to gate delays. In order to avoid this, we must add a gate with $wxz$ also which ensures all adjacent blocks in Kmap are overlapped.
We can also have the sum of products as $xy'z + w'yz + wxy$ which is of course not minimal. But this also has hazard as when $w$ changes from $0$ to $1$, the GATE for $wxy$ should give 1 (from earlier 0) and that of $w'yz$ should give 0 (from earlier 1). But there is a possibility of circuit giving 0 momentarilly (assuming $x=0$). This hazard can be avoided by adding $xyz$ and similarly we need $w'xz$ and $wxz$. Thus we get $ xy'z + w'yz + wxy + xyz + w'xz + wxz$
Now, each of these term will correspond to a GATE when the circuit is implemented. What we require is to avoid the output to change when any of the input literal changes its value (hazard).
B. $wxy, \overline{w}xz,wyz$
Is not correct as $wxy$ is not a minterm for the given function
C. $wx\overline{y} \overline{z}, xz, w\overline{x}yz$

y'z' 
y'z 
yz 
yz' 
w'x' 




w'x 

5 
7 

wx 
12 
13 
15 

wx' 


11 

Here, also there are hazards possible as the middle 4 pairs are separated by 1 bit difference to both $wxy'z'$ as well as $wx'yz$. Could have been avoided by using $wxy'$ instead of $wxy'z'$ and $wyz$ instead of $wx'yz$ which ensures all neighbouring blocks are overlapped.
D. $wx\overline{y}, wyz, wxz, \overline{w}xz, x\overline{y}z, xyz$
These minterms cover all the minterms of $f$ and also, all the neighbouring blocks are overlapping. So, no chance of hazard here and hence is the required answer.