Because these two equations are symmetric, therefore it is sufficient to show the first relation can derive the second one. The other direction can be proved in a similar way.

Assuming the following equation holds $$a_n=\sum_{k=0}^{n}(-1)^k\binom{n+p}{k+p}b_k$$

Setting this to the right side of the second relation yields $$\sum_{k=0}^{n}\sum_{l=0}^{k}(-1)^{k+l}\binom{n+p}{k+p}\binom{k+p}{l+p}b_l$$

Because $$\begin{align*}\binom{n+p}{k+p}\binom{k+p}{l+p}=\ &\frac{(n+p)!}{(k+p)!(n-k)!}\frac{(k+p)!}{(l+p)!(k-l)!}\\ \\=\ &\frac{(n+p)!}{n!}\frac{l!}{(l+p)!}\binom{n}{k}\binom{k}{l} \end{align*}$$

Setting this back to the previous relation will give $$\begin{align*} a_n =\ &\frac{(n+p)!}{n!} \sum_{k=0}^{n}\sum_{l=0}^{k}(-1)^{k+l}\frac{l!}{(l+p)!}\binom{n}{k}\binom{k}{l}b_l \\  \\=\ &\frac{(n+p)!}{n!}\sum_{l=0}^{n}(-1)^l\frac{l!}{(l+p)!}b_l\sum_{k=l}^{n}(-1)^{k}\binom{n}{k}\binom{k}{l}\end{align*}$$

The conclusion of # 4302 gives $$\sum_{k=m}^{n}(-1)^k\binom{n}{k}\binom{k}{m}=0\qquad, m\ne n$$

Therefore the inner sum of the calculation of $a_n$ will only survive when $l=n$. Hence, $$a_n=\frac{(n+p)!}{n!}\cdot (-1)^n\frac{n!}{(n+p)!}b_n(-1)^n=b_n$$