Well, there is a lot of length bashing required!

Dropping a perpendicular (of length 1) from D to AP, and similarly from C to BP, we see that:

AD = cosec 6° DP = cosec 78° BC = cosec 42° CP = cosec 66°

Notice that, for x = 6°, 78°, −42°, −66°, and 30°, sin 5x = ½. We now express sin 5x in terms of sin x.

De Moivre's theorem states that for any real number x and any integer n,

cos nx + i sin nx = (cos x + i sin x)n

Setting n = 5, expanding the right-hand side using the binomial theorem, and equating imaginary parts, we obtain

sin 5x = sin5x − 10 sin3x cos2x + 5 sin x cos4x = sin5x − 10 sin3x (1 − sin2x) + 5 sin x (1 − sin2x)2, since sin2x + cos2x = 1 = 16 sin5x − 20 sin3x + 5 sin x

This result can also be obtained by means of trigonometric identities. Setting s = sin x, it follows that the five distinct real numbers, sin 6°, sin 78°, −sin 42°, −sin 66°, and sin 30° = ½, (1) are roots of the equation 16s^5 − 20s^3 + 5s = ½, or, equivalently, of 32s^5 − 40s^3 + 10^s − 1 = 0. (2) By the Fundamental Theorem of Algebra, (2) has exactly five roots, up to multiplicity, and hence these must be precisely the distinct roots identified in (1).

Since s = ½ is a root of (2), the equation factorizes: (2s − 1)(16s^4 + 8s^3 − 16s^2 − 8s + 1) = 0, yielding the quartic equation whose roots are sin 6°, sin 78°, −sin 42°, and −sin 66°.

As s = 0 is not a root of this quartic equation, we may divide by s4, and, setting t = 1/s, obtain t4 − 8t3 − 16t2 + 8t + 16 = 0, an equation whose roots are cosec 6°, cosec 78°, −cosec 42°, and −cosec 66°.

By Vieta's formulas, the sum of the roots of this equation is 8.

Thus, AD -BC+DP - CP = 8.
Credit should also go to @Sharky Kesa

For some reason the sketch is coming small. Please magnify 200% or more. $AD - BC + PC - PD =\dfrac 1 {Sin6} - \dfrac 1 {Sin42} + \dfrac 1 {Sin66} - \dfrac 1 {Sin78} = \Large\ \ \ \color{#D61F06}{ 8 }$
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Drop perpendicular from P to DC at point Q. Creating Right Triangles PDQ and PCQ.

Sin(78) = 1/DP yielding DP = 1.022

Sin 66) = 1 / PC yielding PC = 1.095

Looking at Triangle PAD and using Law of Sines : 1.022 / Sin (6) = AD / sin (78) which yields AD = 9.564

Look at Triangle PBC and use Law of Sines : 1.095 / sin (42) = BC / sin(66) from which we obtain BC = 1.495

9.564 -1.495 + 1.022 - 1.095 = 7.996 or 8

Use simple geometry, $AD=\csc6^\circ$ , $BC=\csc42^\circ$ , $DP=\csc78^\circ$ , $CP=\csc66^\circ$ ,

hence, $AD-BC+DP-CP=\csc6^\circ-\csc42^\circ+\csc78^\circ-\csc66^\circ=8$