Answer
(a) $m\angle P = 63^{\circ}$
(b) $NP = 13~cm$
Work Step by Step
(a) Since $\angle M \cong \angle Q$, the measure of the angle $m\angle M = 90^{\circ}$
We can see that $m\angle NRQ = 90^{\circ}$ since it is supplementary to $\angle PRN$
Then $m\angle MNR = 90^{\circ}$, since the sum of the four angles in $MNRQ$ is $360^{\circ}$
We can find $m\angle P$:
$m\angle P + m\angle PRN + m\angle RNP = 180^{\circ}$
$m\angle P + m\angle PRN + m\angle RNP = 180^{\circ}$
$m\angle P + m\angle PRN + (m\angle MNP - m\angle MNR) = 180^{\circ}$
$m\angle P + m\angle PRN + (m\angle P+54^{\circ} - m\angle MNR) = 180^{\circ}$
$2m\angle P + m\angle PRN + 54^{\circ} - m\angle MNR = 180^{\circ}$
$2m\angle P + 90^{\circ} + 54^{\circ} - 90^{\circ} = 180^{\circ}$
$2m\angle P + 54^{\circ} = 180^{\circ}$
$2m\angle P = 126^{\circ}$
$m\angle P = \frac{126^{\circ}}{2}$
$m\angle P = 63^{\circ}$
(b) We can find $PR$:
$PR = PQ- QR$
$PR = PQ-MN$
$PR = 20~cm-15~cm$
$PR = 5~cm$
Note that $RN = MQ = 12~cm$
We can find the length of $\overline{NP}$:
$NP = \sqrt{(PR)^2+(RN)^2}$
$NP = \sqrt{(5~cm)^2+(12~cm)^2}$
$NP = \sqrt{25~cm^2+144~cm^2}$
$NP = \sqrt{169~cm^2}$
$NP = 13~cm$
