How to derive the rate laws for three consecutive reactions?
Your first set of four equations look correct. Next you need to integrate the first equation to get [A] vs. time i.e. $\ce{[A]}= \ce{A_0}\exp(-k_1t)$ where $\ce{[A_0]}$ is the value at $t=0$, and substitute [A] into the second equation. Integrate this equation (integrating factor method) and substitute it into the third and so on. It gets messy!  You can obtain $\ce{[P]}$ easily as this is the sum of all other values less the amount you start with.

The general solution for a chain of n reactions in the special case that only the first species A is present at $t=0$ has the form

$$N_n = C_1\exp(-k_1t)~+ C_2\exp(-k_2t) ~+ ...+ C_n\exp(-k_nt) $$

$$C_1= \frac{k_1k_2...k_{n-1}}{(k_2-k_1)(k_3-k_1)...(k_n-k_1)}\ce{[A_0]} $$

$$C_2= \frac{k_1k_2...k_{n-1}}{(k_1-k_2)(k_3-k_2)...(k_n-k_2)}\ce{[A_0]}$$

$$ C_3 = .....$$
etc,  for as many terms as necessary.

good luck!