Multiplying through by $\sqrt{ab}$ we see that we require the largest number $q$ so that
$$\frac{a+b}{\sqrt{ab}} + \frac{\sqrt{ab}}{a+b} p \ge q, \quad \forall a, b > 0.$$
To deal with this, let
$$s = \frac{a+b}{\sqrt{ab}},$$
so that the inequality becomes
$$s + \frac{p}{s} \ge q.$$
This suggests that we should look for the minimum value of the function
$$f(s) = s + \frac{p}{s}.$$
But
$$f(s) - 2\sqrt{p} = \left(\sqrt{s} - \frac{\sqrt{p}}{\sqrt{s}}\right)^2 \ge 0,$$
with equality iff $s = \sqrt{p}$. Hence, $f(s) \ge 2\sqrt{p}$, $\forall s > 0$, i.e., $q \le 2\sqrt{p}$. In other words, for all $a, b > 0$,
$$\frac{1}{a} + \frac{1}{b} + \frac{p}{a+b} \ge \frac{2\sqrt{p}}{\sqrt{ab}}.$$
To prove that $2\sqrt{p}$ is the best constant, we must confirm that there is a pair of positive numbers $a, b$ such that
$$\frac{a+b}{\sqrt{ab}} = \sqrt{p} \iff t + \frac{1}{t} = \sqrt{p}. \quad \text{with } t = \sqrt{\frac{a}{b}}.$$
But $p \ge 4$. Hence $(\sqrt{p} \pm \sqrt{p-4})/2$ are two positive solutions of the quadratic equation $t^2 - \sqrt{p}t + 1 = 0$. Choose $t$ to be any one of these, let $a = t^2, b = 1$ and confirm that
$$\frac{1}{t^2} + 1 + \frac{p}{t^2 + 1} = \frac{2\sqrt{p}}{t}$$
Hence $q = 2\sqrt{p}$.