Answer
$h(p(t)) =\dfrac{4}{1+3 e^{-0.5 t} }$ and $h(p(t)) =\dfrac{4 e}{e+3}$
Work Step by Step
Given $$
h(p)=\frac{4}{p} ; p(t)=1+3 e^{-0.5 t}
$$
Since $h(p(t))$ given by
\begin{align*}
h(p(t))&=\frac{4}{p(t) } \\
&=\frac{4}{1+3 e^{-0.5 t} }
\end{align*}
and $$h(p(t)) =\dfrac{4}{1+3 e^{-1} }=\dfrac{4 e}{e+3}$$
