If $n=3$ the answer is clearly $1$. Now suppose that $n \ge 4$. Consider all the $\binom{n}{2}$ edges between two points in $P$. We colour them blue and red, according to whether they appear in exactly one triangle or not. Counting with multiplicity, we need exactly $3 \binom{n-1}{2}$ edges. On the other hand, any blue edge can be used at most once and a red edge can be used at most $n-2$ times. So we can use at most $\binom{n-1}{2}+(\binom{n}{2}-\binom{n-1}{2})(n-2)=3\binom{n-1}{2}$ edges. So we need to have equality i.e. exactly $\binom{n-1}{2}$ blue edges and exactly $n-1$ red edges, all of them used exactly $n-2$ times. Now consider the possible distribution of the edges. Any point which has one blue edge must have at most one red edge. So any point has either $n-1$ red edges or at most $1$. Clearly, it is not possible that all points have at most one red edge since then their number would be bounded by $\frac{n}{2} <n-1$. So one point must have all $n-1$ edges red and these are all of them. Finally, one can easily check that each such distribution satisfies the conditions. This leaves $n$ choices for $T$ since it is uniquely determined by choosing the one point in $P$ with all edges red.