We'll let you guess.

(I labeled the points wrong so bear with me).

Rotate $\triangle ABC$ about point $A$ by 60 degrees to form a second triangle, as shown. Then $PC=P'C'=3$ , $PB=P'C=5$ , and $AP=AP'=4$ . Because $P'$ is the image of point $P$ after being rotated, $PAP'=60^{\circ}$ . Also notice that $AP=AP'$ , so $\triangle APP'$ must be equilateral. Thus, angle $APP'$ is $60^{\circ}$ . Then, because $CP=3$ , $PP'=4$ , and $CP'=5$ , angle $CPP'$ is right. Finally, the angle $CPA=CPP'+P'PA=90^{\circ}+60^{\circ}=150^{\circ}$ .

Draw a point E such that ADE is an equilateral triangle (look at the picture). Therefore AE = AD = ED = 3 and $\widehat{EAD}$ = $\widehat{EDA}$ = 60 (degrees).

Since ABC is an equilateral triangle, $\widehat{BAC}$ = 60 (degree) and AC = AB. Therefore, $\widehat{BAD}+\widehat{CAD}$ = $\widehat{BAE}+\widehat{BAD}$ . Hence, $\widehat{CAD}=\widehat{BAE}$ .

Now look at the triangles AEB and ADC. They must be equal to each other because AE=AD; $\widehat{CAD}=\widehat{BAE}$ and AC = AB.

Hence EB=CD=5.

The triangle DEB has DE = 3; DB = 4; EB = 5. According to the Pythagorean theorem, DEB must be a right triangle with $\widehat{EDB}$ = 90 (degrees)

In conclusion, $\widehat{ADB}$ = $\widehat{ADE}+\widehat{EDB}$ = 90 + 60 = $\boxed{150}$ (degrees).