= Solution
For $rN>a$ and $p>0$, a nonzero infected <equilibrium> requires $S_*=a/r$. Using $aI_*=pR_*$ and population conservation gives
$$
\boxed{S_*=\frac ar,\qquad I_*=\frac{p}{a+p}\left(N-\frac ar\right),\qquad R_* =\frac{a}{a+p}\left(N-\frac ar\right).}
$$
Eliminate $R=N-S-I$ before assessing <stability>, since perturbations must preserve the total population. The resulting <Jacobian matrix> at this <endemic equilibrium> is
$$
J=\begin{pmatrix}-p-rI_*&-p-a\\rI_*&0\end{pmatrix},\quad \operatorname{tr}J=-\frac{p(p+rN)}{a+p},\quad\det J=p(rN-a).
$$
Its <trace> is negative and its <determinant> positive, so both <eigenvalues> have negative real part: \b[the endemic equilibrium is locally asymptotically stable on the fixed-population plane]. The full three-variable system has a neutral direction corresponding to changing the conserved population.
Writing $\kappa=rN-a>0$, as $p\to0$ the <trace> is $-prN/a+O(p^2)$ and the <determinant> is $p\kappa$. Hence the <discriminant> $(\operatorname{tr}J)^2-4\det J=-4p\kappa+O(p^2)<0$. As $p\to\infty$, the <trace> is $-p-\kappa+O(p^{-1})$ and the <determinant> is again $p\kappa$, giving a positive <discriminant> $p^2+O(p)$. Both quantities are $O(p)$ in both limits, but the <eigenvalues> are \b[complex for sufficiently small $p$ and real for sufficiently large $p$].
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