Let pendulums $P_1$ and $P_2$ have time period $T_1$ and $T_2$ and initial phase $\phi_1$ and $\phi_2$ respectively. For them to have the same phase, $$\sin(\frac{2\pi}{T_1}t+\phi_1) = \sin(\frac{2\pi}{T_2}t+\phi_2)$$$$\implies \frac{2\pi}{T_1}t+\phi_1 = n\pi + (-1)^n(\frac{2\pi}{T_2}t+\phi_2), n\in\mathbb Z$$ From here, you can find solutions for $t$ for even and odd cases of $n$. These solutions of $t$ will be the instants when the two pendulums are in the same position on their phasor diagrams.

Let pendulums $P_1$ and $P_2$ have time period $T_1$ and $T_2$ and initial phase $\phi_1$ and $\phi_2$ respectively. For them to have the same phase, $$\sin\left(\frac{2\pi}{T_1}t+\phi_1\right) = \sin\left(\frac{2\pi}{T_2}t+\phi_2\right)$$ $$\implies \frac{2\pi}{T_1}t+\phi_1 = n\pi + (-1)^n\left(\frac{2\pi}{T_2}t+\phi_2\right), n\in\mathbb Z$$From here, you can find solutions for $t$ for even and odd cases of $n$. These solutions of $t$ will be the instants when the two pendulums are in the same position on their phasor diagrams.