1. Let the side of the original equilateral triangle be $a$.

   Then:

   • Perimeter of the original triangle: $3a$
   • Area of the original triangle: $\frac{\sqrt{3}}{4}a^2$

2. Effect of joining midpoints

   When we join the midpoints of the sides of an equilateral triangle, the new triangle formed is also equilateral, with side length half of the previous one.

   So the sequence of side lengths is: $a,\ \frac a2,\ \frac a4,\ \frac a8,\dots$

3. Sum of perimeters $P$

   Perimeters form a geometric progression: $3a,\ \frac{3a}{2},\ \frac{3a}{4},\dots$

   Hence, $P=3a\left(1+\frac12+\frac14+\cdots\right)$

   Since $1+\frac12+\frac14+\cdots = 2,$ we get $P=3a\cdot 2=6a$

4. Sum of areas $Q$

   Area is proportional to the square of the side, so each new area is $\frac14$ of the previous one.

   Thus areas form the GP: $\frac{\sqrt{3}}{4}a^2,\ \frac{\sqrt{3}}{16}a^2,\ \frac{\sqrt{3}}{64}a^2,\dots$

   Therefore, $Q=\frac{\sqrt{3}}{4}a^2\left(1+\frac14+\frac1{16}+\cdots\right)$

   Now, $1+\frac14+\frac1{16}+\cdots=\frac{1}{1-1/4}=\frac{4}{3}$

   So, $Q=\frac{\sqrt{3}}{4}a^2\cdot \frac43=\frac{\sqrt{3}}{3}a^2$

5. Relate $P^2$ and $Q$

   We have $P=6a \implies P^2=36a^2$

   Also, $Q=\frac{\sqrt{3}}{3}a^2$

   Then, $36\sqrt{3}\,Q=36\sqrt{3}\left(\frac{\sqrt{3}}{3}a^2\right)=36\cdot \frac{3}{3}a^2=36a^2$

   Hence, $P^2=36\sqrt{3}\,Q$

6. Check options

   • A: $P^2=72\sqrt{3}Q$ ❌
   • B: $P^2=36\sqrt{3}Q$ ✅
   • C: $P=36\sqrt{3}Q^2$ ❌
   • D: $P^2=6\sqrt{3}Q$ ❌

Therefore, the correct option is B.