Is it possible to calculate the torsion constant of a rod given the shear modulus, moment of inertia and length?
Maybe not a complete answer to your question, but here are a couple of remarks.
The formula for the period of a torsional pendulum in your original post was wrong. It should be $$T=2\pi\sqrt \frac IK$$ in which $I$ is the moment of inertia of the disc in your diagram and $K$ is the torsional constant of the fibre or rod. The equation is the rotational analogue of the equation for the linear system of a mass on a spring, that is $T=2\pi\sqrt \frac mk$. Moment of inertia is the rotational analogue of mass for a rigid body rotating about a given axis. It takes account of how the body's mass is distributed (since for a rotating body different parts have different linear velocities and accelerations). For a uniform disc of mass $m$ and radius $a$ rotating about its usual rotation axis, the moment of inertia is given by $$I=\tfrac 12 m a^2.$$
The first equation that you quote purports to give the torsional constant in terms of the length and diameter of the rod and the shear modulus of the material of which it is made. The second version of that equation is the one to use, namely $$K=\frac {G\pi d^4}{32 L}.$$ The first version (the one with $I_p$ in it) is dimensionally wrong if $I_p$ is interpreted as an ordinary moment of inertia. In fact $I_p$ is the so-called 'geometrical moment of inertia' of the fibre or rod and it plays a completely different role from the dynamic role of the moment of inertia of the disc. My advice is simply to ignore that version of the $K$ formula, the one with $I_p$ in it.
You now have three usable equations, and there are several relationships you could test and quantities that you could find... Good luck!
You need to be careful with what is called "moment of inertia", as this may refer to different quantities in different contexts. To be solved your problem requires two of these "moments of inertia", which may be quite confusing.
The constant $I_{p}$ appearing in your torsion constant is more rigourously called the second polar moment of area, it characterizes the static deformation of a rod (a wire) in torsion (when applied a torque). The given expression is correct, if you apply a torque $M$ to a rod of length $l$, then you will see that the end section rotates by an angle $\theta$ such that $M=K\theta$. Said differently, if one tries to rotate the rod tip section by $\theta$, the rod will oppose a torque $M=-K\theta$.
Now coming to the second part of your question, namely the relationship between the "moment of inertia" and the period of oscillation: the moment of inertia in question is that of the mass attached to the rod (the red disc in your picture). This quantity which characterizes the dynamics of the rotation of the mass around its axis $z$ is correctly called moment of inertia, let us write it $J_{\Delta z}$. Then applying Newton's second principle for a mass system set in rotation around the $z$ axis : $$J_{\Delta z} \ddot\theta=M=-K\theta $$ This gives you the equation of an harmonic oscillator with period: $$ T=2\pi\sqrt{\frac{J_{Δz}}{K}}$$