Let $n=p+k$, multiply by $z^p$, and sum to compute the generating function:
$$
\begin{align}
\sum_{p=0}^\infty\sum_{m=0}^p\binom{p+k-1-m}{p-m}z^p
&=\sum_{m=0}^\infty\sum_{p=m}^\infty\binom{p+k-1-m}{p-m}z^p\tag1\\
&=\sum_{m=0}^\infty\sum_{p=m}^\infty(-1)^{p-m}\binom{-k}{p-m}z^p\tag2\\
&=\sum_{m=0}^\infty z^m\sum_{p=0}^\infty(-1)^p\binom{-k}pz^p\tag3\\
&=\sum_{m=0}^\infty\frac{z^m}{(1-z)^k}\tag4\\
&=\frac1{(1-z)^{k+1}}\tag5
\end{align}
$$
Explanation:
$(1)$: change order of summation
$(2)$: apply negative binomial coefficients
$(3)$: substitute $p\mapsto p+m$
$(4)$: sum in $p$
$(5)$: sum in $m$
Now use the generating function
$$
\begin{align}
\sum_{m=0}^{n-k}\binom{n-1-m}{n-k-m}
&=\left[z^{n-k}\right]\frac1{(1-z)^{k+1}}\tag6\\
&=(-1)^{n-k}\binom{-k-1}{n-k}\tag7\\
&=\binom{n}{n-k}\tag8\\[6pt]
&=\left[z^{n-k}\right](1+z)^n\tag9\\[6pt]
&=\left[z^{n-k}\right](1+z)^{n-1}\frac1{1-\frac{z}{1+z}}\tag{10}\\
&=\left[z^{n-k}\right](1+z)^{n-1}\sum_{m=0}^\infty\frac{z^m}{(1+z)^m}\tag{11}\\
&=\left[z^{n-k}\right](1+z)^{n-1}\sum_{m=0}^{n-k}\frac{z^m}{(1+z)^m}\tag{12}
\end{align}
$$
Explanation:
$\phantom{1}(6)$: apply $(5)$; i.e. pull out the $p=n-k$ coefficient
$\phantom{1}(7)$: apply the Binomial Theorem
$\phantom{1}(8)$: apply negative binomial coefficients
$\phantom{1}(9)$: apply the Binomial Theorem
$(10)$: $1+z=\left(1-\frac{z}{1+z}\right)^{-1}$
$(11)$: apply the power series for $(1-x)^{-1}$
$(12)$: discard terms with powers of $z$ higher than $n-k$