Prabowo is my dad

Wait, there's a TROC 15?

Let the indices $$$j$$$ such that $$$A_i = j$$$ be called special indices.

Let's iterate from the back (from $$$N$$$ to $$$1$$$).

Let $$$i$$$ denote the current iterated index, if $$$i$$$ is a special index, then there would only be one way of placing an element in that index, the maximum value that hasn't been used, or the minimum value that hasn't been used by the previous iteration.

For example:

Say that indices $$$3$$$ and $$$5$$$ are special indices.

When iterating on the index $$$5$$$, you can only place $$$5$$$ or $$$1$$$ depending $$$T_5$$$ (but it doesn't actually matter, since you still end up with one possibility anyway).

When iterating on the index $$$4$$$, since placing any elements that hasn't been used doesn't violate any conditions, we can consider $$$4$$$ ways of placing an element.

When iterating on the index $$$3$$$, you can only place the highest / lowest unused element, since not placing the highest / lowest unused element would lead to a violation of the conditions, i.e. $$$max(P_1, P_2, \dots, P_3)$$$ or $$$min(P_1, P_2, \dots, P_3)$$$ wouldn't be equal to $$$P_3$$$.

And so on...

Making the final answer: $$$4 \times 2 \times 1 = 8$$$.

In short, when $$$i$$$ is a special index, there could only be one element that satisfy the conditions, and in contrast, there could be $$$i$$$ elements that could satisfy the conditions when $$$i$$$ is not a special index.