Answer
(a). $P=P_0e^{-\frac{h}{k}},$
(b). $P=56.472$
Work Step by Step
$\ln \left(\frac{P}{P_0}\right)=-\frac{h}{k}$
$k=7, P_0=100kpa$
(a).
$e^{\ln {\frac{P}{P_0}}}=e^{-\frac{h}{k}},$
$\frac{P}{P_0}=e^{-\frac{h}{k}},$
$P=P_0e^{-\frac{h}{k}},$
(b).$h=4,$
$P=100e^{-\frac{4}{7}},$
$P=56.472$