In the rays $DA$ and $BA$ we take a point $E$ and $Z$, respectively, such that $AC = AE = AZ$.

Since $\widehat{DEC} = \frac{\widehat{DAC}}{2} = 18^\circ = \widehat{CBD}$, the quadrilateral $DEBC$ is cyclic.

Similarly, the quadrilateral $CBZD$ is cyclic, because
$$\widehat{AZC} = \frac{\widehat{BAC}}{2} = 36^\circ = \widehat{BDC}.$$

Figure 8

Figure 9

Therefore the pentagon $BCDZE$ is inscribed in the circle $K(A, AC)$. It gives $AC = AD$ and $\widehat{ACD} = \widehat{ADC} = \frac{180^\circ - 36^\circ}{2} = 72^\circ$, which gives $\widehat{ADP} = 36^\circ$ and $\widehat{APD} = 108^\circ$.

Alternative solution:
See the figure in the right. If $C_1$ is the symmetric point of $D$ with respect to the line $BC$, then $ABC_1D$ is inscribable. Since $\widehat{BAC_1} = 72^\circ$ and $\widehat{BDD_1} = 36^\circ$ we have that $DD_1$ is bisector of the angle $BDC_1$ and hence $C_1$ is the incenter of the triangle $BDC_1$. Therefore we have
$$\widehat{DC_1A} = \widehat{AC_1B} = 36^\circ \quad \text{and} \quad \widehat{DPA} = \frac{72^\circ + 144^\circ}{2} = 108^\circ.$$