If cos P

(b) $60^{\circ}=\frac{\pi}{3}$

$\cos P=\frac{1}{7}, \quad \cos Q=\frac{13}{14}$

Therefore, $\sin P=\sqrt{1-\frac{1}{49}}=\frac{4 \sqrt{3}}{7}$ and $\sin Q=\sqrt{1-\frac{169}{196}}=\frac{3 \sqrt{3}}{14}$

Hence, $\tan P=4 \sqrt{3}, \quad \tan Q=\frac{3 \sqrt{3}}{13}$

$\cos (P-Q)=\cos P \cos Q+\sin P \sin Q$

$=\frac{1}{7} \times \frac{13}{14}+\frac{4 \sqrt{3}}{7} \times \frac{3 \sqrt{3}}{14}$

$=\frac{13+36}{98}$

$=\frac{49}{98}$

$\therefore \cos (P-Q)=\frac{1}{2}$

$\Rightarrow P-Q=\cos ^{-1} \frac{1}{2}$

$\Rightarrow P-Q=60^{\circ}$

Hence, the correct answer is option B.