Linearization around the equilibrium point for a system of differential equations

Suppose we have a system of differential equations

$$ \frac{dN}{dt} = \mu N \left(1-\frac{N}{K}\right) + N \int_{0}^{\infty}p(a,t)da $$ $$ \frac{\partial p}{\partial t} + \frac{\partial p}{\partial a} = k p(a,t). $$

Linearizing it around $(N=K,p=0)$, I am getting the system

$$ \frac{dN}{dt} = h - \mu N + N \int_{0}^{\infty}p(a,t)da $$ $$ \frac{\partial p}{\partial t} + \frac{\partial p}{\partial a} = k p(a,t),$$ where $h = \mu K.$

The reason I am getting this system is because I am considering $N \int_{0}^{\infty}p(a,t)da$ as linear term. Is this linearisation right or the correct system is

$$ \frac{dN}{dt} = h - \mu N $$ $$ \frac{\partial p}{\partial t} + \frac{\partial p}{\partial a} = k p(a,t),$$ where $h = \mu K.$

I agree with the linearisation of the second equation. For the first, you are close but not quite correct. To find the linearisation around $N=K$, $p=0$, substitute $N=K+\varepsilon \tilde N$ and $p = \varepsilon \tilde p$ into your equations and throw out everything that isn't order $\varepsilon$. I got: \begin{align*} \varepsilon \frac{d \tilde N}{dt}=\frac{d N}{dt}&=\mu N \bigg ( 1- \frac N K \bigg ) + N \int_0^\infty p(a,t) \, da \\ &=-\mu (K+\varepsilon \tilde N) \varepsilon \bigg ( \frac {\tilde N} K \bigg ) + \varepsilon (K+\varepsilon \tilde N) \int_0^\infty \tilde p(a,t) \, da \\ &= \bigg ( -\mu \tilde N +K \int_0^\infty \tilde p (a,t) \, da\bigg) \varepsilon +o(\varepsilon). \end{align*} Thus, the linearised equation is (dropping the tildes): $$ \frac{d N}{dt} = -\mu N +K \int_0^\infty p (a,t) \, da .$$

Edit: Answer to question in the comments. Let $N^\ast$, $p^\ast$ be steady state solutions to your system. Replacing $p$ with $p^\ast + \varepsilon p$ the equation for $p$ becomes: \begin{align*} kp^\ast+\varepsilon kp=k (p^\ast+\varepsilon p) &= \frac{\partial p^\ast } {\partial t}+ \varepsilon \frac{\partial p } {\partial t} + \frac{\partial p^\ast}{\partial a}+ \varepsilon \frac{\partial p}{\partial a} \\ &= \frac{\partial p^\ast}{\partial a}+ \varepsilon \bigg ( \frac{\partial p } {\partial t} + \frac{\partial p}{\partial a} \bigg ) \end{align*} which implies the linearised equation is $$kp= \frac{\partial p } {\partial t} + \frac{\partial p}{\partial a}$$ (not surprising since the original equation was linear). For the equation for $N$ replace $N$ and $p$ with $N^\ast +\varepsilon N$ and $p^\ast + \varepsilon p$ respectively: \begin{align*} \varepsilon \frac{dN}{dt} &= \frac{d (N^\ast+\varepsilon N)}{dt} \\ &= \mu (N^\ast+\varepsilon N)\bigg ( 1 - \frac {N^\ast+\varepsilon N} K \bigg ) +(N^\ast+\varepsilon N) \int_0^\infty (p^\ast(a)+\varepsilon p(a,t)) \, da\\ &=\mu N^\ast\bigg ( 1 - \frac {N^\ast} K \bigg ) +N^\ast \int_0^\infty p^\ast\, da \\ &\qquad + \varepsilon \bigg [ N^\ast \int_0^\infty p\,da +N \int_0^\infty p^\ast \, da + \mu N \bigg (1-\frac{2N^\ast}K \bigg )\bigg ] \\ &\qquad + \varepsilon^2 \bigg [N \int_0^\infty p \, da -\frac{\mu N^2}{K}\bigg ] \\ &=\varepsilon \bigg [ N^\ast \int_0^\infty p\,da +N \int_0^\infty p^\ast \, da + \mu N \bigg (1-\frac{2N^\ast}K \bigg )\bigg ] \\ &\qquad + \varepsilon^2 \bigg [N \int_0^\infty p \, da -\frac{\mu N^2}{K}\bigg ] \end{align*} using that $(N^\ast,p^\ast)$ is the steady state solution. Thus, the linearised equation is: $$\frac{dN}{dt} = N^\ast \int_0^\infty p\,da +N \int_0^\infty p^\ast \, da + \mu N \bigg (1-\frac{2N^\ast}K \bigg ).$$