Answer: $0.2Y_{t-1} + 0.6Y_{t-4} - 0.12Y_{t-5} + \epsilon_t$

This question requires the application of the properties of the Lag operator. Recall that: $$LY_t = Y_{t-1}$$ And that the lag operator has multiplicative property. So, $$ \begin{aligned} (1 - 0.2L)(1 - 0.6L^4)Y_t &= (1 - 0.6L^4 - 0.2L + 0.12L^5)Y_t \\ &= Y_t - 0.6Y_tL^4 - 0.2Y_tL + 0.12Y_tL^5 = \epsilon_t \\ &= Y_t - 0.6Y_{t-4} - 0.2Y_{t-1} + 0.12Y_{t-5} = \epsilon_t \end{aligned} $$ Rearranging we get: $$Y_t = 0.2Y_{t-1} + 0.6Y_{t-4} - 0.12Y_{t-5} + \epsilon_t$$ Therefore, option A is correct.

Author: Nikitesh Somanthe