The general $SARIMA(p,d,q)(P,D,Q)_m$ process $X_t$ is the solution of the following equation
$$\Phi(B^m)\phi(B)\nabla_m^D\nabla^d X_t=\Theta(B^m)\theta(B)Z_t,$$
where $Z_t$ is the white noise process. $\nabla_mX_t=X_{t}-X_{t-m}$, $\nabla X_t=X_{t}-X_{t-1}$,
\begin{align}
\Phi(B^m)&=1-\Phi_1B^m -\dots-\Phi_PB^{PM}\\
\phi(B)&=1-\phi_1B-\dots-\phi_p B^{p}\\
\Theta(B^m)&=1-\Theta_1 B^m-\dots-\Theta_Q B^{Qm}\\
\theta(B&)=1-\theta_1B-\dots-\theta_qB^q
\end{align}
and $B^nX_t=X_{t-n}$.
Now take your example of $ARIMA(1,0,3)(1,2,0)_5$. This means that
\begin{align}
\Phi(B^m)&=\Phi(B^5)=1-\Phi_1B^5\\
\phi(B)&=1-\phi_1B\\
\Theta(B^m)&=\Theta(B^5)=1\\
\theta(B&)=1-\theta_1B-\theta_2B^2-\theta_3B^3
\end{align}
So in your case $X_t$ must satisfy the equation:
\begin{align}
(1-\Phi_1B^5)(1-\phi_1B)\nabla_5^2X_t=(1-\theta_1B-\theta_2B^2-\theta_3B^3)Z_t
\end{align}
which we can rewrite as
\begin{align}
(1-\phi_1 B - \Phi_1 B^5 +\phi_1\Phi_1 B^{6})\nabla_5^2X_t=Z_t-\theta_1Z_{t-1}-\theta_2Z_{t-2}-\theta_3Z_{t-3}.
\end{align}
Now $\nabla^2_5X_t=\nabla_5(X_t-X_{t-5})=(X_t-2X_{t-5}+X_{t-10})$ and I leave the last step for the reader to complete.