Label  ΔABC.


The given information in ΔABC is

$\begin{array}{rcl}\angle C& =& 112°\\ a& =& 5.6 \mathrm{cm}\\ c& =& 8.2 \mathrm{cm}\end{array}$

$\angle A\left(\angle H\right)$ is across from side a = 5.6 cm ($\angle A$ and side a are a matching pair.)
$\angle C=112°$ and is across from side c = 8.2 cm ($\angle C$ and side c are a matching pair.)


Since $\angle A=\angle H, \angle H$ is equal to 39°.

Label ΔABC.


The given information in ΔABC is

$\begin{array}{rcl}\angle C& =& 71°\\ b& =& 2.8 \mathrm{m}\\ c& =& 6.5 \mathrm{m}\end{array}$

$\angle B$ is across from side b = 2.8 m. ($\angle B$ and side b are a matching pair.)
$\angle C=71°$ is across from side c = 6.5 m. ($\angle C$ and side c are a matching pair.)
$\angle A$, or $\angle P$, is across from side a.

Since side a is not given, $\angle A$ (or $\angle P$) cannot be yet be calculated. However, the sine law can be used to calculate $\angle B$.

$\begin{array}{rcl}\frac{\sin B}{b}& =& \frac{\sin {\displaystyle C}}{c}\\ \frac{{\displaystyle \sin B}}{{\displaystyle 2.8}}& =& \frac{{\displaystyle \sin 71°}}{{\displaystyle 6.5}}\\ \frac{{\displaystyle \sin B}}{{\displaystyle \cancel{2.8}}}\times \cancel{2.8}& =& \frac{{\displaystyle \sin 71°}}{{\displaystyle 6.5}}\times 2.8\\ \sin B& =& \frac{{\displaystyle \sin 71°}}{{\displaystyle 6.5}}\times 2.8\\ & =& 0.4073\\ \angle B& =& {\sin }^{-1}0.473\\ & =& 24°\end{array}$

Recall that the sum of the angles in a triangle is 180°. Use $\angle A+\angle B+\angle C=180°$ to find $\angle A$.

$\begin{array}{rcl}\angle A+\angle B+\angle C& =& 180°\\ \angle A+24°+71°& =& 180°\\ \angle A+95°& =& 180°\\ \angle A& =& 180°-95°\\ & =& 85°\end{array}$

Since $\angle A=\angle P, \angle P$ is equal to 85°.