A body rotating around a fixed axis relative to another rotating body A question in my book goes as follows:

A planet P revolves around the Sun in a circular orbit, with the Sun at the center, which is co-planar with and concentric to the circular orbit of Earth E around the Sun. P and E revolve in the same direction. The times required for the revolution of P and E around the Sun are $T_\mathrm{P}$ and $T_\mathrm{E}$. Let $T_\mathrm{S}$ be the time required for P to make one revolution around the Sun relative to E: show that $\dfrac{1}{T_\mathrm{S}} = \dfrac{1}{T_\mathrm{E}} - \dfrac{1}{T_\mathrm{P}}.$ Assume $T_\mathrm{P} > T_\mathrm{E}$

What does 'the time required by planet P to revolve around the Sun relative to the Earth' mean?
 A: My interpretation of this statement is that your reference frame is on Earth, $T_S$ is how long it takes for another planet $P$ to come back to the same spot relative to the Earth.
Let's take an example: $P$ is the planet Mars
The period of the Earth $T_P=365.25$ days. The period of Mars $T_P=687$ days (a quick Google search). By the formula, $\frac{1}{T_S}=\frac{1}{365.25}-\frac{1}{687}\Rightarrow T_S=779.8$ days
This is the time required by Mars to revolve around the Sun relative to Earth, for example, the time between two closest approaches of Mars.
According to NASA, the time between two closest approaches of Mars is $26$ months $\approx 793$ days (if we assume an average month has $30.5$ days).
Proof:

Source of picture
We know that, looking at the Earth's orbit, that $\theta=(T_S-T_E)(\frac{2\pi}{T_E})$ ($T_S-T_E$ is the time it takes for the Earth to moves through the angle and we know that the angle is $T\frac{2\pi}{T_E}$, where $T_E$ is the sidereal period, the "real" period of the Earth, $1$ year).
Looking at the planet's orbit, that $\theta=T_S\frac{2\pi}{T_P}$.
Equate the two expressions for $\theta$, and we get
$(T_S-T_E)(\frac{2\pi}{T_E})=T_S\frac{2\pi}{T_P}$. Divide by $2\pi$, and bring the $T_S$ down, we get $\frac{T_S-T_E}{T_ET_S}=\frac{1}{T_P}$, and we get $\frac{1}{T_P}=\frac{1}{T_E}-\frac{1}{T_S}$.
Proof came from here
A: What does the time required by planet P to revolve around the Sun relative to the Earth mean?
That's your basic question.
To which the answer is
It means the time required by P (which moves slower than E because itá distance to the center of the circle is bigger than the distance of E to the center, which means it has more distance to cover and on top of that, P's velocity is smaller: $\frac{mv^2}{r}=G\frac{mM}{r^2}$, so $v=G\sqrt{\frac{M}{r}}$ ) to make one revolution relative to the time it takes E to make (obviously) more than one revolutions.
Depending on both the distances of E ($r_E$) and P ($r_P$) you can calculate the difference in their rotation times.
