 Solution

Neither asymmetric purchase outcome is a Nash equilibrium for any permitted $P>0,n>0$. At $(b,nb)$, X's chosen purchase requires $P\le 2t/3$, but Y's chosen nonpurchase, against X buying, requires $P\ge t$. These requirements are incompatible because $2t/3<t$.

At $(nb,b)$, Y's chosen purchase requires $P\le t/3$, whereas X's nonpurchase against a buying Y requires $P\ge t$. Again the inequalities are incompatible. Thus every Nash equilibrium in pure strategies has both players taking the same action; the conclusion is stronger than merely observing symmetry in their available actions, since their original market shares differ.