Define $P_\phi := e^{-iP\phi}$, where $P$ is a Pauli matrix with some overall phase factor and $\phi\in[0,2\pi)$. It is claimed (see Page 1 of this paper) that if $P'P = -PP'$ i.e. we have two anticommuting Pauli matrices $P, P'$, then
$$P_{\frac{\pi}{4}}P'_{\phi} = (iPP')_{\phi}P_{\frac{\pi}{4}}.$$
Note that $iPP'$ is also a Pauli matrix with some phase factor and we can use the notation introduced. How can one show this identity?
EDIT:
My mistake with the example previously here. As correctly remarked in the answer, I missed a sign and it should be
$$P_{\frac{\pi}{4}}P'_{\phi} = (-iPP')_{\phi}P_{\frac{\pi}{4}}.$$
Your relation is a special case of the general formula $$e^{-i\vec{\alpha} \cdot \vec{\sigma}/2}e^{-i \vec{\beta} \cdot \sigma/2} e^{i \vec{\alpha}\cdot \vec{\sigma}/2}=e^{-i \left(R(\vec{\alpha}\right) \vec{\beta})\cdot\vec{\sigma}/2}, \tag{1} \label{1} $$ where $$\vec{\alpha} \cdot \vec{\sigma} = \sum\limits_{k=1}^3 \alpha_k \sigma_k\equiv \alpha_k \sigma_k, \quad \vec{\alpha} \in \mathbb{R}^3 \tag{2} \label{2}$$ is a linear combination of the Pauli matrices $\sigma_1, \sigma_2, \sigma_3$ (summation convention used in $\alpha_k \sigma_k$). $R(\vec{\alpha}) \in \rm SO(3)$ is a rotation matrix defined by $$ \qquad R(\vec{\alpha})\vec{\beta}= (\cos \alpha) \, \vec{\beta}+(1-\cos \alpha ) \vec{n} \left(\vec{n} \cdot \vec{\beta}\right) +(\sin \alpha) \,\vec{n} \times \vec{\beta}, \quad \vec{\alpha} = \alpha \vec{n}, \; |\vec{n}|=1. \tag{3} \label{3}$$ Eq. \eqref{1} can be shown by using the basic relations $$[\sigma_k, \sigma_\ell ]= 2i \epsilon_{k \ell m} \sigma_m \quad \text{and} \quad \{\sigma_k, \sigma_\ell\}= 2 \delta_{k \ell} \mathbb{I}_2, \tag{4} \label{4}$$ implying $$e^{-i \vec{\alpha} \cdot \vec{\sigma}/2} = \mathbb{I}_2\cos\frac{\alpha}{2}-i \vec{n} \cdot \vec{\sigma} \sin \frac{\alpha}{2} \tag{5}$$ and $$ e^{-i \vec{\alpha}\cdot \vec{\sigma}/2} \,\vec{\beta}\cdot \vec{\sigma} \,e^{i \vec{\alpha} \cdot \vec{\sigma}/2}= \left(R(\vec{\alpha}) \vec{\beta}\right) \cdot \vec{\sigma}, \tag{6} \label{6}$$ proving \eqref{1} by exponentiation.
As a side remark, eq. \eqref{6} establishes the relation $\rm SU(2)/_{Z_2} \simeq SO(3)$ between the groups $\rm SU(2)$ and $\rm SO(3)$. Note that the $\rm SO(3)$ element $R(\vec{\alpha})$ can be obtained from the $\rm SU(2)$ element $U(\vec{\alpha})=e^{-i \vec{\alpha} \cdot \vec{\sigma}/2}$ using $$R_{k \ell} = \frac{1}{2}{\rm Tr}\left(\sigma_k U\sigma_\ell U^\dagger \right), \tag{7} \label{7}$$ showing that the the $\rm SU(2)$ elements $\pm U$ are mapped to the same $\rm SO(3)$ element $R$.
Coming back to your specific problem, we observe that the relations in \eqref{4} imply $$ (\vec{\alpha} \cdot \vec{\sigma}) (\vec{\beta} \cdot \vec{\sigma})= \vec{\alpha} \cdot \vec{\beta} \;\mathbb{I}_2+i (\vec{\alpha} \times \vec{\beta}) \cdot \vec{\sigma}, \tag{8} \label{8}$$ so $\{\vec{\alpha} \cdot \vec{\sigma}, \vec{\beta} \cdot \vec{\sigma} \}=0$ is equivalent to $\vec{\alpha} \cdot \vec{\beta}=0$. In this case, \eqref{3} simplifies to $$R(\vec{\alpha}) \vec{\beta} = (\cos{\alpha}) \, \vec{\beta} + (\sin \alpha) \, \vec{n} \times \vec{\beta}, \qquad \vec{\alpha} \perp \vec{\beta}. \tag{9} \label{9}$$ Specializing further to $\alpha=\pi/2$, we have $$ R(\vec{n} \pi/2) \vec{\beta} = \vec{n} \times \vec{\beta}. \tag{10} \label{10}$$ For this choice of $\vec{\alpha}$, eq. \eqref{1} leads to $$e^{-i \vec{n}\cdot \vec{\sigma} \pi/4} e^{-i \vec{\beta} \cdot \vec{\sigma}/2}= e^{-i (\vec{n} \times \vec{\beta}) \cdot \vec{\sigma}/2} e^{-i \vec{n} \cdot\vec{\sigma} \pi/4} = e^{-(\vec{n}\cdot \vec{\sigma}) (\vec{\beta} \cdot \vec{\sigma})}e^{-i \vec{n} \cdot \vec{\sigma} \pi/4}, \qquad \vec{n} \perp \vec{\beta}.\tag{11} \label{11}$$ Translating into your favourite (but somewhat unconventional) notation, $$ P = \vec{n} \cdot \vec{\sigma}, \quad P^\prime = \vec{n}^\prime \cdot \vec{\sigma}, \quad P_{\pi/4} =e^{-i\vec{n} \cdot \vec{\sigma} \pi/4}, \quad P^\prime_\phi=e^{-i \vec{n}^\prime \cdot \vec{\sigma} \phi}, \tag{12} \label{12} $$ we have $$P P^\prime = i (\vec{n} \times \vec{n}^\prime) \cdot \vec{\sigma} \quad \Rightarrow \quad (-i P P^\prime)_\phi = e^{-i (\vec{n} \times \vec{n}^\prime) \cdot \vec{\sigma} \phi} \tag{13}$$ and eq. \eqref{11} with $\beta = 2 \phi$ becomes $$P_{\pi/4} P^\prime _\phi =(-i \,P P^\prime)_\phi P_{\pi/4} \tag{14} \label{14}$$ with a sign discrepancy compared to the formula shown in your question.