Calculate the angle x In the figure, AB is diameter, PM = MH and PN = NB.
If PEB arc = 108 °, calculate "x"

Drawing to obtain the right triangle in P: $\Delta APB $.  $\hat{A}$ as $\hat{B}$  are angles inscribed on the circumference, they are equal to half the arc you see. Therefore,  $\hat{A} = 54^o$ , $\hat{B} = 36^o$ and  note also that $\overline {MN}$ t is  average base $\Delta HPB $.
If $\hat{B} = 36^o$. $\hat{H} = 90^o$ then  $H\hat{P}B = 54^o$

In the right triangle, the middle base is parallel to the base of the original triangle. Therefore, the triangle $\Delta PMN$ is $\hat{M}$ rectangle is $P \hat{N} M = 36^o$ and In addition $R \hat{Q}N$ and $O \hat{Q}M$ are opposed by the vertex. From this, it is easy to conclude that $\Delta OMQ$ $\simeq$ $\Delta NRQ$
Therefore,  in the triangle $\Delta OMQ$ we obtain that $ x + 90^o + 36^o = 180^o \Rightarrow x = 54^o $
how to demonstrate the similarity of $\Delta OMQ$ $ and \Delta NRQ$ ?
 A: 
The sketch shown in $\mathrm{Fig.\space 1}$ is an extended version of the diagrams you have included in your problem statement. We denote the sought angle $\measuredangle RQN$ as $\alpha$.  Let $F$ be the center of the semicircle $BPA$ and $\measuredangle NRO=\phi$.
As you have mentioned in your incomplete answer, $\measuredangle PAB$ and $\measuredangle ABP$ can be easily determined as $54^o $ and $36^o$ respectively, because $\measuredangle PFB =108^o$. It is also given that $M$ and $N$ are the midpoints of the segments $PH$ and $PB$ respectively. Therefore $MN$ is parallel to $AB$, which makes $\measuredangle MNR = 36^o$.
To solve this seemingly difficult problem we have to bring Ceva’s theorem in to play. That is why we have extended the line $NM$ to intersect $PA$ at $D$. This makes $D$ the midpoint of $PA$. As you see below, solving of the problem now becomes as easy as eating a piece of cake. Consider the triangle $PAN$, in which three cevians $ND$, $PO$, and $AR$ are concurrent at $M$. Applying Ceva’s theorem to $\triangle PAN$, we obtain,
$$\frac{PD}{DA}\frac{AO}{ON}\frac{NR}{RP}=1. \tag{1}$$
Since $PD=DA$, equation (1) can be simplified to get,
$$\frac{AO}{ON}=\frac{RP}{NR}. \tag{2}$$
Equation (2) implies that $PA$ is parallel to $RO$. Therefore, we have $\phi = \measuredangle NRO = \measuredangle BPA = 90^o$. Using triangle $RQN$, we obtain,
$$\alpha = \measuredangle RQN =180^o-\measuredangle QNR\space\space –\space\phi =180^o – 36^o – 90^o = 54^o.$$
