{P1, P2, ..., Pn}
Pi holds Xi instances.
Pi can request additional Yi instances.
Given two process p & q such that their additional requests are zero.
Yp = Yq = 0
{Yk | 1 ≤ k ≤ n, k ≠ p, k ≠ q} means that out of 'n' processes, we are left with (n-2) process (except p&q), i.e., Yk indicates additional request of all the processes (n-2) except p & q.
For p & q to complete first, accordingly
Xp + Xq < Min {Yk}
Option A is correct.
There are exactly two process p and q which do not need any additional instances of resources.
So, p and q will complete their execution and will release Xp and Xq instances of resources.
Now to guarantee that no other process apart from p and q can complete execution, the no. of instances of resources available must be less than the minimum no. of instances of resources required by any other process, i.e.,
Xp + Xq < Min {Yk | 1 ≤ k ≤ n, k ≠ p, k ≠ q}
For example, given by @srestha in one of the comment
P1,P2,P3,P1,P2,P3
Now,
P1 holding 1 instance of resource R
P2 holding 2 instances of resource R
P3 holding 3 instances of resource R
And R totally has 6 resources
Now P1 need some additional instances
and P2,P3,P2,P3 do'not need any additional instances of R , So, after releasing 5 instances of R is free.
Now P1 must need a minimum more than 5 instances
So, A