If $a = b = 1$ the parameter $p > 0$ has to satisfy:
$$\begin{cases} 2\sqrt{p+1} \ge p+1, \\ 2 \ge \sqrt{p+1}, \\ p \le 3. \end{cases}$$
We show that for $p \in (0, 3)$ the inequality holds for any real $a$ and $b$. If $p \in (0, 1)$ the inequality holds trivially:
$$\sqrt{a^2 + p b^2} > a, \quad \sqrt{b^2 + p a^2} > b \quad \text{and} \quad (p-1)\sqrt{ab} \le 0.$$
Let further be $p \in (1, 3)$. The left hand side, LHS, of the inequality can be understood as the sum of the lengths of vectors $(a, b\sqrt{p})$ and $(b, a\sqrt{p}) \in \mathbb{R}^2$. According to the triangle inequality then
$$\begin{aligned} LHS &= \sqrt{a^2 + p b^2} + \sqrt{b^2 + p a^2} = |(a, b\sqrt{p})| + |(b, a\sqrt{p})| \\ &\ge |(a+b, (a+b)\sqrt{p})| = (a+b)\sqrt{1+p}. \end{aligned} \quad (1)$$
For the RHS we have (with the help of AM-GM inequality)
$$\text{RHS} = a+b+(p-1)\sqrt{ab} \le a+b+(p-1)\frac{a+b}{2} = \frac{(p+1)(a+b)}{2}.$$
Now $LHS \ge RHS$ evidently, because even stronger inequality
$$(a+b)\sqrt{p+1} \ge \frac{(p+1)(a+b)}{2}$$
is equivalent to $\sqrt{p+1} \le 2$, which is obviously satisfied for any $p \in (1, 3)$.