Lyla and Isabelle run on a circular track both starting at point $P$. Lyla runs at a constant speed in the clockwise direction. Isabelle also runs in the clockwise direction at a constant speed $25 \%$ faster than Lyla. Lyla starts running first and Isabelle starts running when Lyla has completed one third of one lap. When Isabelle passes Lyla for the fifth time, how many times has Lyla returned to point $P$ ?
We will imagine that the track has been broken into 12 equal units.

With this convention, Isabelle will begin to run once Lyla has ran 4 units, and Isabelle will always run 5 units in the amount of time it takes Lyla to run 4 units.

When Lyla has ran a total of 4 units, Isabelle has ran 0 units. At this time, Isabelle starts to run, so when Lyla has ran $4+4=8$ units, Isabelle will have ran 5 units.

When Lyla has ran $8+4=12$ units, Isabelle will have ran $5+5=10$ units.

When Lyla has ran $12+4=16$ units, Isabelle will have ran $10+5=15$ units.

When Lyla has ran $16+4=20$ units, Isabelle will have ran $15+5=20$ units.

At this point, Isabelle passes Lyla for the first time.

Since Isabelle runs faster than Lyla, Isabelle will always have ran farther than Lyla from this point on.

In the time it takes Lyla to run 12 units, Isabelle will run 15 units.

Continuing the calculation above, when Lyla has ran $20+12=32$ units, Isabelle will have ran $20+15=35$ units.

When Lyla has ran $32+12=44$ units, Isabelle will have ran $35+15=50$ units.

When Lyla has ran $44+12=56$ units, Isabelle will have ran $50+15=65$ units.

When Lyla has ran $56+12=68$ units, Isabelle will have ran $65+15=80$ units.

At this point, Isabelle is exactly 12 units, or one lap, ahead of Lyla.

Similarly, at any time when Isabelle passes Lyla, the difference between the number of units they have ran will be a multiple of 12 .

This means the 5th time Isabelle passes Lyla will occur when Isabelle has gained an additional $3 \times 12=36$ units on Lyla.

Isabelle gains one unit on Lyla for every 4 units Lyla runs, which means that Isabelle will gain 36 units on Lyla when Lyla runs $4 \times 36=144$ units.

Therefore, when Isabelle passes Lyla for the 5th time, Lyla has ran a total of $68+144=212$ units.

There are 12 units in a lap, so this means Lyla will have ran $\frac{212}{12}=17 \frac{8}{12}$ laps.

When Isabelle passes Lyla for the 5th time, Lyla will have completed 17 laps.

ANSWER: 17