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The opposite angles of a parallelogram are congruent.

In $â–\pm JQKS$, angles KSJ and KQJ are opposite angles. Therefore, $∠KSJâ‰\dots ∠KQJ$.

From the given figure, it can be observed that $\stackrel{¯}{SK}âˆ\yen \stackrel{¯}{JQ}$and $\stackrel{¯}{KQ}$is a transversal then $∠2$and $∠KQJ$are alternate interior angles such that, $∠2â‰\dots ∠KQJ$.

From the given figure, it can be observed that $\stackrel{¯}{SK}âˆ\yen \stackrel{¯}{JQ}$and $\stackrel{¯}{SJ}$is a transversal then $∠1$and $∠KSJ$are alternate interior angles such that, $∠1â‰\dots ∠KSJ$.

As $∠1â‰\dots ∠KSJ$, $∠2â‰\dots ∠KQJ$and $∠KSJâ‰\dots ∠KQJ$it implies that $∠1â‰\dots ∠2$.

The opposite sides of a parallelogram are congruent.

In $â–\pm JQKS$, sides $\stackrel{¯}{SJ}$and $\stackrel{¯}{KQ}$are opposite sides. Therefore, $\stackrel{¯}{SJ}â‰\dots \stackrel{¯}{KQ}$.

As $\stackrel{¯}{PJ}â‰\dots \stackrel{¯}{RK}$, $∠1â‰\dots ∠2$and $\stackrel{¯}{SJ}â‰\dots \stackrel{¯}{KQ}$then by SAS postulate $\mathrm{Δ}SPJâ‰\dots \mathrm{Δ}KRQ$.

As $\mathrm{Δ}SPJâ‰\dots \mathrm{Δ}KRQ$then by corresponding parts of congruent triangles, $∠Pâ‰\dots ∠R$.

Hence it is proved that $∠Pâ‰\dots ∠R$.