Answer
$y'=e^{p}+\dfrac{3}{2}p^{1/2}e^{p}+pe^{p}+p^{3/2}e^{p}$
Work Step by Step
$y=e^{p}(p+p\sqrt{p})$
Rewrite the function like this:
$y=e^{p}[p+(p)(p^{1/2})]=e^{p}(p+p^{3/2})$
Differentiate using the product rule
$y'=(e^{p})(p+p^{3/2})'+(p+p^{3/2})(e^{p})'=...$
$...=(e^{p})(1+\dfrac{3}{2}p^{1/2})+(p+p^{3/2})(e^{p})=...$
Evaluate the products and simplify:
$...=e^{p}+\dfrac{3}{2}p^{1/2}e^{p}+pe^{p}+p^{3/2}e^{p}$
