What is θ in Bent's rule equation (%s - character)?
TL;DR: $\theta$ is the angle between any two equivalent bonds and using  Coulson's theorem for anything beyond the second period is out of its scope.

First things first, Coulson's theorem (the generalization of said formula) cannot be applied in cases of spndm. Therefore using $\ce{PCl5}$ as a comparison is not the right idea. Also, the formula that you've stated is only when finding the bond angle between two equivalent bonds as would be proved below.
This equation is supposedly a simplification of Coulson's theorem that states,
$$1 +\lambda_1\lambda_2\cos(\theta) = 0$$
Here, $\lambda_1$ stands for the hybridization index of one bonded orbital, $\lambda_2$ denotes the hybridization index of the second bonded orbital and $\theta$ signifies the bond angle between the two orbitals. However, this rule as said above can only be applied to spn hybridisation compounds where $\sqrt{n}$ is the hybridisation index. Its formal proof is given in Wikipedia if you are curious and want to find out.
The simplified idea behind Coulson's theorem (as I understand it), is as follows.
Each set of equivalent bonds is made up of a linear combination between s and p orbitals. This is what hybridization is. Given any pair of bonds, we can find the bond angle between the two through their hybridization indexes ($\lambda_1$ and $\lambda_2$) and vice versa. Let's take the example of $\ce{CH4}$,
$\ce{CH4}$ has four equivalent $\ce{C-H}$ bonds and we know that the bond angle is $109.5^\circ$and so according to Coulson's theorem, the hybridisation of the $\ce{C=H}$ bond can be found to be,
$$1+\lambda_\ce{C-H}\cdot\lambda_\ce{C-H}\cos(\theta_\ce{H-C-H}) = 0 $$
$$1 +\lambda_\ce{C-H}^2\cos(\theta_\ce{H-C-H})  = 0 $$
\begin{align}
\therefore \lambda_\ce{C-H}^2 &amp;= \frac{-1}{\cos(109.5^\circ)} \\
&amp;\approx 3
\end{align}
Therefore, the $\ce{C-H}$ bonds are sp3 in nature.

Proof of $\cos \theta = \frac{s}{s-1}$
From Coulson's theorem, we have,
$$1 +\lambda_1\lambda_2\cos(\theta) = 0$$
Now, if both bonds are equivalent in nature, we get $\lambda_1 = \lambda_2 = \lambda$, therefore,
$$1 + \lambda^2\cos(\theta) = 0$$
This implies,
$$\cos \theta = \frac{-1}{\lambda^2}$$
Now for such a molecule the hybridisation would be sp$\lambda^2$ and so percentage s-character would be
$$\text{%s character} = \frac{1}{1 + \lambda^2}$$
Solving for $\lambda^2$, we get:
$$\lambda^2 = \frac{1-s}{s}$$
Substituting this in the equation for $\cos \theta$, we get:
$$\cos \theta = \frac{s}{s-1}$$
Note: An assumption that I took here is that the bonds are equivalent. If this is in fact true, then this rule can only be used to find the bond angle between any two equivalent bonds and hence in the case of $\ce{PCl5}$, only the axial-axial ($\ce{P-Cl}$) bonds or equatorial-equatorial $\ce{(P-Cl)}$ bonds can be compared. θ then is the angle between any two equivalent bonds.