(1)

外接円を持つので対角の和は\(\pi\)となるので、
\begin{align*} \cos B & =\frac{a^{2}+b^{2}-p^{2}}{2ab}\\ & =\frac{a^{2}+b^{2}-\left(c^{2}+d^{2}-2cd\cos D\right)}{2ab}\\ & =\frac{a^{2}+b^{2}-c^{2}-d^{2}+2cd\cos\left(B+D-B\right)}{2ab}\\ & =\frac{a^{2}+b^{2}-c^{2}-d^{2}+2cd\left\{ \cos\left(B+D\right)\cos\left(-B\right)-\sin\left(B+D\right)\sin\left(-B\right)\right\} }{2ab}\\ & =\frac{a^{2}+b^{2}-c^{2}-d^{2}-2cd\cos\left(B\right)}{2ab}\\ & =\frac{ab}{ab+cd}\frac{a^{2}+b^{2}-c^{2}-d^{2}+2cd\sin\left(B+D\right)\sin\left(B\right)}{2ab}\\ & =\frac{a^{2}+b^{2}-c^{2}-d^{2}}{2\left(ab+cd\right)} \end{align*} となり、与式は成り立つ。

(2)

(1)より、
\begin{align*} p & =\sqrt{a^{2}+b^{2}-2ab\cos B}\\ & =\sqrt{a^{2}+b^{2}-2ab\frac{a^{2}+b^{2}-c^{2}-d^{2}}{2\left(ab+cd\right)}}\\ & =\sqrt{a^{2}+b^{2}-ab\frac{a^{2}+b^{2}-c^{2}-d^{2}}{\left(ab+cd\right)}}\\ & =\sqrt{cd\frac{a^{2}+b^{2}}{ab+cd}+ab\frac{c^{2}+d^{2}}{ab+cd}}\\ & =\sqrt{\frac{cd\left(a^{2}+b^{2}\right)+ab\left(c^{2}+d^{2}\right)}{ab+cd}} \end{align*} となり、与式は成り立つ。