Linearization around the equilibrium point for a system of differential equations \frac{dN}{d

Step 1
I agree with the linearisation of the second equation. For the first, you are close but not quite correct. To find the linearisation around ${\displaystyle N=K,p=0}$, substitute ${\displaystyle N=K+\epsilon \tilde{N}}$ and ${\displaystyle p=\epsilon \tilde{p}}$ into your equations and throw out everything that isn't order ${\displaystyle \epsilon }$. I got:
$\begin{array}{rl}\epsilon \frac{d\tilde{N}}{dt}=\frac{dN}{dt}& =\mu N(1-\frac{N}{K})+N{\int }_0^{\mathrm{\infty }}p(a,t)da\\ & =-\mu (K+\epsilon \tilde{N})\epsilon \left(\frac{\tilde{N}}{K}\right)+\epsilon (K+\epsilon \tilde{N}){\int }_0^{\mathrm{\infty }}\tilde{p}(a,t)da\\ & =(-\mu \tilde{N}+K{\int }_0^{\mathrm{\infty }}\tilde{p}(a,t\left)da\right)\epsilon +o\left(\epsilon \right).\end{array}$
Step 2
Thus, the linearised equation is (dropping the tildes):
${\displaystyle \frac{dN}{ dt }=-\mu N+K{\int }_0^{\mathrm{\infty }}p(a,t),da}$.
Edit: Answer to question in the comments. Let ${\displaystyle N\cdot ,p\cdot }$ be steady state solutions to your system. Replacing p with ${\displaystyle p^{\ast }+\epsilon p}$ the equation for p becomes:
$\begin{array}{rl}k{p}^{\ast }+\epsilon kp=k(p^{\ast }+\epsilon p)& =\frac{\mathrm{\partial }{p}^{\ast }}{\mathrm{\partial }t}+\epsilon \frac{\mathrm{\partial }p}{\mathrm{\partial }t}+\frac{\mathrm{\partial }{p}^{\ast }}{\mathrm{\partial }a}+\epsilon \frac{\mathrm{\partial }p}{\mathrm{\partial }a}\\ & =\frac{\mathrm{\partial }{p}^{\ast }}{\mathrm{\partial }a}+\epsilon (\frac{\mathrm{\partial }p}{\mathrm{\partial }t}+\frac{\mathrm{\partial }p}{\mathrm{\partial }a})\end{array}$
which implies the linearised equation is
${\displaystyle kp=\frac{\partial p}{\partial t}+\frac{\partial p}{\partial a}}$
(not surprising since the original equation was linear). For the equation for $N$ replace $N$ and p with ${\displaystyle N^{\ast }+\epsilon N}$ and ${\displaystyle p^{\ast }+\epsilon p}$ respectively:
Step 3
$\begin{array}{rl}\epsilon \frac{dN}{dt}& =\frac{d(N^{\ast }+\epsilon N)}{dt}\\ & =\mu (N^{\ast }+\epsilon N)(1-\frac{N^{\ast }+\epsilon N}{K})+(N^{\ast }+\epsilon N){\int }_0^{\mathrm{\infty }}\left(p^{\ast }\right(a)+\epsilon p(a,t\left)\right)da\\ & =\mu N^{\ast }(1-\frac{N^{\ast }}{K})+N^{\ast }{\int }_0^{\mathrm{\infty }}{p}^{\ast }da\\ & +\epsilon [N^{\ast }{\int }_0^{\mathrm{\infty }}pda+N{\int }_0^{\mathrm{\infty }}{p}^{\ast }da+\mu N(1-\frac{2N^{\ast }}{K}\left)\right]\\ & +{\epsilon }^2[N{\int }_0^{\mathrm{\infty }}pda-\frac{\mu N^2}{K}]\\ & =\epsilon [N^{\ast }{\int }_0^{\mathrm{\infty }}pda+N{\int }_0^{\mathrm{\infty }}{p}^{\ast }da+\mu N(1-\frac{2N^{\ast }}{K}\left)\right]\\ & +{\epsilon }^2[N{\int }_0^{\mathrm{\infty }}pda-\frac{\mu N^2}{K}]\end{array}$
using that ${\displaystyle (N^{\ast },p^{\ast })}$ is the steady state solution. Thus, the linearised equation is:
${\displaystyle \frac{dN}{ dt }=N^{\ast }{\int }_0^{\mathrm{\infty }}p,da+N{\int }_0^{\mathrm{\infty }}{p}^{\ast },da+\mu N(1-\frac{2N^{\ast }}{K}).}$