I recommend using FindInstance
to obtain a single solution for each allowed value of $a$. The following uses generating functions to answer your questions about counting the number of solutions with and without multiplicity.
Rewrite in terms of variables ${u,v,w}$, where $x=u$, $y=u+v$, and $z=u+v+w$. The constraints on ${x,y,z}$ become $u\ge1$, $v\ge0$, and $w\ge0$. The original equation $x+y+z=p+a$ becomes $3u+2v+w=p+a$.
The series in dummy variable $s$ corresponding to $3u$ where $u\ge1$ is $s^3/(1-s^3)=s^3+s^6+s^9+...$.
The series corresponding to $2v$ where $v\ge0$ is $1/(1-s^2)=1+s^2+s^4+s^6+...$.
The series corresponding to $w$ where $w\ge0$ is $1/(1-s)=1+s^1+s^2+s^3+...$.
The coefficient of $s^{p+a}$ in the product of these three series give the number of integer solutions to $x+y+z=p+a$.
Series[s^3/(1 - s^3)*(1/(1 - s^2))*(1/(1 - s)), {s, 0, 15}]
s^3 + s^4 + 2 s^5 + 3 s^6 + 4 s^7 + 5 s^8 + 7 s^9 + 8 s^10 +
10 s^11 + 12 s^12 + 14 s^13 + 16 s^14 + 19 s^15
For example, the above series shows that for parameter $p=5$, when $a=-5$ to $a=5$ there are ${0,0,0,1,1,2,3,4,5,7,8}$ solutions, respectively.
The coefficients of $s^0$, $s^1$, and $s^2$ are always zero because of the $3u$ term and the constraint that $u\ge1$.
Given parameter $p$, the total number of solutions for all allowed $a$ between $a=-p$ and $a=p$ including multiplicity is given by the following generating function. One additional term $1/(1-s)$ is included in the product. The coefficient of $s^{2p}$ of this series gives the total number of solutions to the equation, including multiplicity.
Series[s^3/(1 - s^3)*(1/(1 - s^2))*(1/(1 - s))*(1/(1 - s)), {s, 0, 15}]
s^3 + 2 s^4 + 4 s^5 + 7 s^6 + 11 s^7 + 16 s^8 + 23 s^9 + 31 s^10 + 41 s^11 + 53 s^12
Given parameter $p$, all $a$ between $a=-p+3$ and $a=p$ produce at least one solution to $x+y+z=p+a$. Therefore, the count of solutions without multiplicity is $2(p-1)$.
// DeleteDuplicatesBy[First] // Column
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