Suppose the 8 seats are labeled 

A,B,C,D,E,F,G,H 

with

1.  seat A opposite seat E,
2.  seat B opposite seat F,
3.  seat C opposite seat G,
4.  seat D opposite seat H

The two people can choose a pair of seats in any of C(8,2) = 28 ways.
They can choose a pair of opposite seats in only 4 ways.

So the probability is 4 out of 28 ways or  which reduces to .

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Explanation:

That's these four choices of ways to sit opposite each other:  

{A,E} {B,F} (C,G}, {D,H}

out of these 28 ways to sit opposite or not:

{A,B), {A,C}, {A,D}, {A,E}, {A,F}, {A,G}, {A,H}
{B,C), {B,D}, {B,E}, {B,F}, {B,G}, {B,H}, {C,D}
{C,E), {C,F}, {C,G}, {C,H}, {D,E}, {D,F}, {D,G}
{D,H), {E,F}, {E,G}, {E,H}, {F,G}, {F,H}, {G,H}

Edwin