I think the solution above is wrong. Here is mine. Suppose the contrary, let $p=ab+cd$. Clearly, $p>a$, hence $a^{-1},b^{-1},c^{-1},d^{-1}$ modulo $p$ exist. Now, $ab+cd\equiv 0\pmod{p}$ implies $a\equiv -cd/b\pmod{p}$. Inserting this, we find that \[ \frac{c^2d^2}{b^2}-\frac{c^2d}{b}-c^2\equiv b^2+bd-d^2\pmod{p}\implies c^2d^2-c^2bd-c^2b^2\equiv b^4+b^3d-b^2d^2\pmod{p}. \]Collecting the terms, we find, \[ \left(b^2+c^2\right)\left(b^2+bd-d^2\right)\equiv 0\pmod{p}. \]Hence, there are two cases to investigate. Case 1. Assume $p\mid b^2+c^2$. Then, using \[ \left(a^2+d^2\right)\left(b^2+c^2\right)=\left(ab+cd\right)^2+\left(ac-bd\right)^2 \]Since $p\mid b^2+c^2$ and $ab+cd=p$, it follows $p\mid ac-bd$. Now, note that using $a>b>c>d\ge 1$, we find $0<ac-bd<ab+cd$, thus $p\nmid ac-bd$. A contradiction is reached. Case 2. Let $p\mid b^2-d^2+bd$. Then, $p\mid a^2-c^2+ac$. It is immediately seen that $b^2-d^2+bd<2p$, hence $p=b^2-d^2+bd=a^2-c^2+ac$. Using $ab+cd=b^2-d^2+bd$, we find $b(a-b)=d(b-c-d)$. Since $a-b>0$, it follows $b>c+d$. Now, using $ab+cd=a^2-c^2+ac$, we obtain $a(a-b)=c(c+d-a)$. This yields $c+d-a>0$, that is $c+d>a$. Combining, we find $b>c+d>a$, but we are given $a>b$. Again, a contradiction. Hence, $ab+cd$ is not a prime, completing the proof.