Case 1:
Consider the following example, If x < q, then we will only search the left subtree of q, and no element can be greater than q there, so if anything greater than x comes later in the left subtree that must be less than q. Now consider this relation recursively, so whenever we encounter an element that is greater than x, that must be less than the previous element greater than x we encountered. So all the elements greater than x we encounter during probing, that must be in a decreasing sorted order in a valid probe sequence.
Case 2:
Consider the following example, If x > q, then we will only search the right subtree of q, and no element can be less than q there, so if anything less than x comes later in the right subtree that must be greater than q. Now consider this relation recursively, so whenever we encounter an element that is less than x, that must be greater than the previous element less than x we encountered. So all the elements less than x we encounter during probing, that must be in a increasing sorted order in a valid probe sequence.
Conclusion
So our conclusion is that elements greater than x must be in a decreasing order and element less than x must be in an increasing order in a valid probe sequence searching for x.
Now apply the conclusion to the given options, we can clearly see that (b) has 60 coming after 50 and (e) has 18 coming after 27.