We have $y'(x)=2px^{p-1}$. Now we do something that is not absoutely necessary, but that may give additional clarity. We divide our problem into $2$ cases: (i) $p \ge 1$; (ii) $0
0$.
Case (i): We are interested in $\displaystyle\int_0^b \frac{\sqrt{1+4p^2x^{2p-2}}}{\sqrt{2}x^{p/2}}\,dx$.
The numerator is bounded above in our interval. If $p<2$, then $\displaystyle\int_0^b \frac{dx}{x^{p/2}}$ converges, so our integral converges.
What about if $p \ge 2$? Then our integrand is bigger than $\dfrac{1}{\sqrt{2}x^{p/2}}$, so the integral diverges.
Case (ii): The numerator can be rewritten as $\dfrac{1}{x^{1-p}}\sqrt{x^{2-2p}+4p^2}$. The part $\sqrt{x^{2-2p} +4p^2}$ is bounded above, since $0
In summary, our integral converges when $0
Comment: Here is a slicker way of doing things uniformly for all $p<2$. Note that
$$1+4p^2x^{2p-2}<(1+2px^{p-1})^2.$$
So our integrand is less than
$$\frac{1}{x^{p/2}} +\frac{2p}{x^{1-p/2}}.$$
If $p<2$, then each of $\displaystyle\int\frac{dx}{x^{p/2}}$ and $\displaystyle\int\frac{2p\,dx}{x^{1-p/2}}$ converges.
The downside of this approach is that it is probably less natural than the first approach we gave.