Let's understand Option wise
In the solution, I am assuming x and y both belong to P so I am not writing repeatedly
Option A).
$∀_{x}$ ( xRx → $∃_{y}$ . xRy) which is equivalent to $∀_{x}$ x $∃_{y}$ .( xRx → xRy)
it translates to “ for every value of x there is some y such that if x is reflexive or xRx then xRy”.
this is true because we don’t have any counter-example for x, for example for element “f” fRf is true so xRx → xRy is also true as there is no mention of x != y.
Option B). $∃_{x}$ $∀_{y}$ xRy
it translates to “There is some x for all y in a domain such that xRy”.
as we need some witness in this case and we have no witness in the domain P.
Option C).$∃_{x}$ ( xRx → $∀_{y}$ . xRy)
(I also made a mistake in this option and didn’t mark it but now I realize my silly mistake.)
This is true as it translates to “ There is some x for all y such that if xRx then xRy”.
now here implication is given and it is also true when the antecedent part is false.
so we just need to find such x which is not related to itself we have such x in domain P which is “element C”.
so C is our witness here, C doesn’t relate to itself and C doesn’t relate to all elements in the domain too.
so F→ F = T.
Option D).
$∀_{x}$ $∃_{y}$ xRy
it translates to “ for every x there is some y such that xRy”
This is also true as we don’t have any counter-example to make it false, for element “f” fRf as there is no mention of x != y.
{Most trickiest option is Option C I must say.}
