Assertion (A): In parallelogram ABCD, PD bisects ∠ADC and PC bisects angle DCB, then ∠DPC = 90°.

Reason (R): ∠PDC = $\dfrac{1}{2}$ x ∠ADC

∠PCD = $\dfrac{1}{2}$ x ∠BCD

∠PDC + ∠PCD = $\dfrac{1}{2}$ x (∠ADC + ∠BCD)

1. A is true, but R is false.

2. A is false, but R is true.

3. Both A and R are true, and R is the correct reason for A.

4. Both A and R are true, and R is the incorrect reason for A.

We know that consecutive angles of a parallelogram are supplementary.

ABCD is a parallelogram.

∴ ∠ADC + ∠BCD = 180° …………………(1)

PD bisects ∠ADC.

⇒ ∠PDC = $\dfrac{∠ADC}{2}$ ……………(2)

PC bisects ∠BCD.

⇒ ∠PCD = $\dfrac{∠BCD}{2}$ ……………(3)

Adding equations (2) and (3), we get :

⇒ ∠PDC + ∠PCD = $\dfrac{∠ADC}{2}$ + $\dfrac{∠BCD}{2} = \dfrac{1}{2}$ (∠ADC + ∠BCD)

= $\dfrac{1}{2}$ x 180°

= 90°.

In ΔPCD, according to angle sum property,

⇒ ∠PDC + ∠PCD + ∠DPC = 180°

⇒ 90° + ∠DPC = 180°

⇒ ∠DPC = 180° - 90°

⇒ ∠DPC = 90°

∴ Both A and R are true, and R is the correct reason for A.

Hence, option 3 is the correct option.