Correct Answer: 50$^\circ$

Solution :
Given, that two circles touch each other externally at P. AB is the common tangent to the circles at points A and B respectively. Draw a tangent at point P which meets AB at C.
In a triangle PAC, CA = CP [lengths of the tangents from an external point C].
So, $\angle$CAP = $\angle$APC = 40$^\circ$
Similarly in triangle PBC, CB = CP
So, $\angle$CPB = $\angle$ABP
Now in the triangle APB,
$\angle$APB + $\angle$ABP + $\angle$PAB = 180$^\circ$
⇒ ( 40$^\circ$ + $\angle$CPB ) + $\angle$ABP + 40$^\circ$ = 180$^\circ$
⇒ ( 40$^\circ$ + $\angle$ABP ) + $\angle$ABP + 40$^\circ$ = 180$^\circ$
$\therefore \angle$ABP = 50$^\circ$
Hence, the correct answer is 50$^\circ$.