In your question, you don't define $P(\theta)$ or $R_z(\theta)$. I'm going to assume:
$$
P(\theta)=\left(\begin{array}{cc} 1 & 0 \\ 0 & e^{i\theta} \end{array}\right)\qquad R_z(\theta)=\left(\begin{array}e^{-i\theta} & 0 \\ 0 & e^{i\theta} \end{array}\right).
$$
In this case, you simply have that
$$
R_z(\theta)=P(2\theta)e^{-i\theta}\equiv P(2\theta),
$$
the point being that global phases are irrelevant. However, the difference is important when you look at the controlled-gates. Let's say we can create either controlled-$P$ or controlled-$R_z$. We can create the other via the identity

The extra $P(-\theta)$ is the gate that compensates for the extra phase.
To make controlled-$R_z$, the trick is to notice the identities $R_z(\theta_1)R_z(\theta_2)=R_z(\theta_1+\theta_2)$ and $XR_z(\theta)X=R_z(-\theta)$, where we will replace the Xs by controlled-not so that the change in rotation angle only happens if the control qubit is 1. Hence, we have
