Converting transfer function that is a sum of unusual rational polynomials to finite difference equation
The expression:
$$
H(z) = cz^{-m} + \frac{b_0 z^{-1} + b_1z^{-2} + \dots + b_{2m}z^{-2m}}{1 + a z^{-1}} + \frac{q_0 z^{-1} + q_1z^{-2} + \dots + q_{2m}z^{-2m}}{1 + p z^{-1}}
$$

Can be rewritten as:
$$
H_1(z)=cz^{-m} \\
H_2(z)=\frac{b_0 z^{-1} + b_1z^{-2} + \dots + b_{2m}z^{-2m}}{1 + a z^{-1}}\\
H_3(z)=\frac{q_0 z^{-1} + q_1z^{-2} + \dots + q_{2m}z^{-2m}}{1 + p z^{-1}}\\
H(z) = H_1(z)+ H_2(z)+ H_3(z)
$$

Hence:
$$
y_1(t)=cu(t-m)\\
y_2(t)=b_0 u(t-1) + \dots + b_{2m}u(t-2m) - a y_2(t-1)\\
y_3(t)=q_0 u(t-1) + \dots + q_{2m}u(t-2m) - p y_3(t-1)\\
y(t) = y_1(t)+y_2(t)+y_3(t)
$$

which is a multivariable finite difference equation.

The standard transfer function is obtained simply by multiplying and simplifying terms:
$$
H(z)=\frac{c(1+(a+p)z^{-1}+apz^{-2})z^{-m} 
+ b_0(1+pz^{-1})z^{-1} + \dots + b_{2m}(1+pz^{-1})z^{-2m}  
+ q_0(1+az^{-1})z^{-1} + \dots + q_{2m}(1+az^{-1})z^{-2m} }{(1+az^{-1})(1+pz^{-1})} 
$$
which leads to the univariable finite difference equation, which can be simplified further.....:
$$
y(t)=c(1+(a+p)z^{-1}+apz^{-2})y(t-m) 
+ b_0(1+pz^{-1})z^{-1} + \dots + b_{2m}(1+pz^{-1})u(t-2m)  
+ q_0(1+az^{-1})z^{-1} + \dots + q_{2m}(1+az^{-1})u(t-2m) - (a+p)z^{-1}-apz^{-2}
$$