Lets examine the first sum. I can't seem to find a closed form, but there is something very nice with the generating series. They are simple, and symmetrical with respect to the variables $p$ and $k$.
Result: Your sum is the $k^{th}$ coefficient of $\frac{(1+x)^{p}}{\left(1-x\right)^{p+1}},$ and also the $p^{th}$ coefficient of $\frac{(1+x)^{k}}{\left(1-x\right)^{k+1}}.$
The Generating Series for the variable $p$
Consider
$F(x)=\sum_{p=0}^{\infty}\sum_{i=0}^{p}\binom{k}{i}\binom{k+p-i}{p-i}x^{p}.$
Changing the order of summation, this becomes
$F(x)=\sum_{i=0}^{\infty}\binom{k}{i}\sum_{p=i}^{\infty}\binom{k+p-i}{p-i}x^{p},$
and then shifting the second sum we have
$F(x)=\sum_{i=0}^{\infty}\binom{k}{i}x^{i}\sum_{p=0}^{\infty}\binom{k+p}{p}x^{p}.$ Since the rightmost sum is $\frac{1}{(1-x)^{k+1}}$ we see that the generating series is $F(x)=\frac{1}{(1-x)^{k+1}}\sum_{i=0}^{\infty}\binom{k}{i}x^{i}=\frac{\left(1+x\right)^{k}}{(1-x)^{k+1}}$
by the binomial theorem.
The Generating Series for the variable $k$:
Lets consider the other generating series with respect to the variable $k$. Let
$G(x)=\sum_{k=0}^{\infty}\sum_{i=0}^{p}\binom{k}{i}\binom{k+p-i}{p-i}x^{k}.$
Then
$G(x)=\sum_{i=0}^{p}\sum_{k=i}^{\infty}\binom{k}{i}\binom{k+p-i}{p-i}x^{k}=\sum_{i=0}^{p}x^{i}\sum_{k=0}^{\infty}\binom{k+i}{i}\binom{k+p}{p-i}x^{k}.$
Splitting up the binomial coefficients into factorials, this is
$=\sum_{i=0}^{p}x^{i}\sum_{k=0}^{\infty}\frac{(k+i)!}{k!i!}\frac{(k+p)!}{(k+i)!(p-i)!}x^{k}=\sum_{i=0}^{p}\frac{x^{i}p!}{i!\left(p-i\right)!}\sum_{k=0}^{\infty}\frac{\left(k+p\right)!}{k!p!}x^{k}.$
Consequently,
$G(x)=\frac{(1+x)^{p}}{\left(1-x\right)^{p+1}}.$
Comments: I am not sure why the generating series has this symmetry. Perhaps you can use this property to tell you more about the sum/generating series.
Hope that helps,