If $ b$ is odd, then it is easy to check that $ p$ is not an integer. So, $ b$ is even. Let $ b = 2c$ for some positive integer $ c$. Then, $ \left(\frac{2p}{c}\right)^{2}=\frac{a-c}{a+c}$ Let $ \frac{n}{m}=\frac{2p}{c}$ with $ gcd(m,n) = 1$ and let $ g = gcd(a+c,a-c)$ Then, $ a+c = gm^{2}, a-c = gn^{2}$ Or, $ 2a = g(m^{2}+n^{2}), 2c = g(m^{2}-n^{2})$ So, $ 4p =\frac{2cn}{m}=\frac{ng(m^{2}-n^{2})}{m}$ Since $ gcd(m,n) = 1 , gcd(m,m^{2}-n^{2}) = 1$ , we have $ m | g$ , say $ g = mk$ for some positive integer $ k$. Thus, $ 4p = nk(m^{2}-n^{2})$ Note that $ p$ is a prime and so we can cosider the following two cases. Case 1. $ m , n$ are both odd. Since $ 4p = nk(m^{2}-n^{2})\equiv 0 (\mod 8)$, we obtain $ p = 2$ and hence $ nk\times\frac{m^{2}-n^{2}}{8}= 1$ i.e., $ nk =\frac{m^{2}-n^{2}}{8}= 1$ Thus, $ n = k = 1, m = g = 3, a = 15,c = 12, b = 24$ i.e., $ (p,a,b) = (2,15,24)$ Case 2. $ m , n$ have different parity. i.e., $ (m,n) = (odd,even)$ or $ (even, odd)$ Since $ m^{2}-n^{2}$ is odd and $ 2c = g(m^{2}-n^{2})$ , we have $ 2 | g$ , say $ g = 2s$ for some positive integer $ s$. So, $ 2p =\frac{cn}{m}=\frac{ns(m^{2}-n^{2})}{m}$ Since $ gcd(m,n) = 1 , gcd(m,m^{2}-n^{2}) = 1$ , we have $ m | s$ , say $ s = mt$ for some positive integer $ t$. Thus, $ 2p = nt(m+n)(m-n)$ $ \cdots$ (*) Note that the lefthand side of (*) has at most 2 nontrivial factors $ 2, p$. ① $ n = 1$ $ 2p = t(m+1)(m-1)$ since $ m-1 < m+1$ , $ m-1 = 1$ or $ 2$. If $ m-1 = 2$ , then $ p = 4t$ , a contradiction. So, $ m-1 = 1$ and hence $ 2p = 3t$ , $ p = 3, t = 2$ Thus, $ g = 8, a = 20, c = 12, b = 24$. i.e., $ (p,a,b) = (3,20,24)$ ② $ n > 1$ Since $ m+n > n > 1$, we have $ m+n = p$ . So, $ n = 2 , t = m-n = 1$ Thus, $ m = 3, p = 5 , g = 6 , a = 39, c = 15, b = 30$. i.e., $ (p,a,b) = (5,39,30)$ Therefore, the maximum possible value of $ p$ is $ 5$.

I approached this in a slightly different manner. If $b$ is odd then $2a-b$ is odd so henceforth $\frac{b}{4}\sqrt{\frac{2a-b}{2a+b}}$ cannot be an integer. We consider two cases: $b=4k$, or $b=2m$ where $m$ is odd. Case 1: $b=4k$ \begin{eqnarray*} p=k\sqrt{\frac{2a-4k}{2a+4k}}=k\sqrt{\frac{a-2k}{a+2k}}\\ \left(\frac{p}{k}\right)^2=\frac{a-2k}{a+2k} \\ p^2a+2p^2k=k^2a-2k^3 \\ a(k^2-p^2)=2k(k^2+p^2) \\ a=\frac{2k(k^2+p^2)}{k^2-p^2} \end{eqnarray*} We know that $a$ must be an integer. Consider the cases $p|k$ and $p\nmid k$. The first case gives us $k=mp$. Therefore \[a=\frac{2mp(m^2p^2+p^2)}{m^2p^2-p^2}=\frac{2mp(m^2+1)}{m^2-1}\] We have $\gcd(m, m^2-1)=1$ therefore we must have \begin{align} \frac{2p(m^2+1)}{m^2-1}=2p\left(1+\frac{2}{m^2-1}\right)\in \mathbb{Z}\\ \implies \frac{4p}{m^2-1}\in \mathbb{Z} \end{align} When $m$ is even we must have $(m^2-1)\mid p\implies p=m^2-1=(m-1)(m+1)$ since $p$ is a prime. Therefore $m-1=1$ which gives us $(p,m)=(3,2)$. When $m$ is odd we must have $\frac{m^2-1}{4}\mid p$ therefore $p=\frac{m^2-1}{2}$ or $p=\frac{m^2-1}{4}$. The first case renders no solutions while the second renders the solution $p=2, m=3$. $p=2$ gives the solution $(a,b,p)=(15, 24, 2)$ and $p=3$ gives us $(a,b,p)=(20,24,3)$. Now when $p\nmid k$ we must have $a=\frac{2k(k^2+p^2)}{k^2-p^2}$ be an integer. However $\gcd(k, k^2-p^2)=1$ therefore we must have $\frac{2(k^2+p^2)}{k^2-p^2}=2+\frac{4p^2}{k^2-p^2}$ be an integer. However $\gcd(p^2, k^2-p^2)=1$ therefore we must have $\frac{4}{k^2-p^2}$ be an integer. However, this gives $\begin{cases} k^2-p^2=1 \\ k^2-p^2=2 \\ k^2-p^2=4 \end{cases}$ of which gives no positive solution for $p$. Case 2: $b=2m$ where $m$ is odd \begin{eqnarray*} p=\frac{m}{2}\sqrt{\frac{2a-2m}{2a+2m}} \\ p=\frac{m}{2}\sqrt{\frac{a-m}{a+m}} \\ \left(\frac{2p}{m}\right)^2=\frac{a-m}{a+m} \\ 4p^2a+4p^2m=am^2-m^3 \\ a(m^2-4p^2)=m(4p^2+m^2) \\ a=\frac{m(4p^2+m^2)}{m^2-4p^2}=\frac{m(m^2+n^2)}{m^2-n^2} \end{eqnarray*} Where $n=2p$. Since $\gcd(m,2)=1$ we have two cases to consider: $p\mid m$ and $\gcd(m,n)=1$. When $\gcd(m,n)=1$ however we have $\gcd(m, m^2-n^2)=1$ therefore we must have $\frac{m^2+n^2}{m^2-n^2}=1+\frac{2n^2}{m^2-n^2}$ be an integer. $\gcd(n^2, m^2-n^2)=1$ therefore we must have $\begin{cases} m^2-n^2=1 \\ m^2-n^2=2 \end{cases}$ which gives no valid solutions. Therefore $p\mid m$. Let $m=pk$. We arrive at \[a=\frac{pk(4p^2+p^2k^2)}{p^2(k^2-4)}=\frac{pk(4+k^2)}{k^2-4}\] Keeping in mind the fact that $k$ must be odd we arrive at $\gcd(k(k^2+4), k^2-4)=1$. Therefore we must have $\frac{p}{k^2-4}$ be an integer or henceforth $k^2-4\mid p$. Since $p$ is prime we arrive at $p=k^2-4$. This is only solved when $k=3$ which gives $p=5$. This gives the solution pair $(a,b,p)=(39,30,5)$. In conclusion the answer is $p=5$.