Take a point $C'$ on the line $AR$ in such a way that the points $A, R, C'$ lie on the line in this order and $RC' = 1$ is satisfied. From the hypothesis of the problem, it then follows that $RC' = PC, RQ = PQ$ are satisfied, and furthermore, we obtain the fact that the triangles $C'RQ$ and $CPQ$ are congruent since
$$\begin{aligned} \angle C'RQ &= 360^\circ - (\angle RQP + \angle QPB + \angle PBR) = 360^\circ - (90^\circ + \angle QPB + 90^\circ) \\ &= 180^\circ - \angle QPB = \angle CPQ. \end{aligned}$$
Consequently, we have $\angle AQC' = \angle AQR + \angle RQC' = \angle AQR + \angle PQC = 90^\circ$. Let $AR = AQ = 2x$. Then, we have $QC = QC' = x$, and by applying the Pythagorean Theorem to the triangle $AQC'$, we get $AC' = \sqrt{AQ^2 + (QC')^2} = \sqrt{5}x$. From $1 = RC' - AR = (\sqrt{5} - 2)x$ we have $x = \frac{1}{\sqrt{5}-2}$, and therefore,
$$AR = 2x = \frac{2}{\sqrt{5}-2} = 2\sqrt{5} + 4.$$