Because p is prime $d=p-a,c=p-b, p>4\to a+b+c+d=2p$ 2 of members $a^{2013},b^{2013},c^{2013},d^{2013}$ - odd. Therefore $2|A=a^{2013}+b^{2013}+c^{2013}+d^{2013}$ $p|a^{2013}+d^{2013}=a^{2013}+(p-a)^{2013}$ and $p|b^{2013}+c^{2013}$, therefore $p|A$ and $a+b+c+d=2p|A$.

My solution: In $\mathbb{Z}/p\mathbb{Z}$ we have $x^4=y^4$ for some non-zero distinct integers $x,y$ if and only if $\left(x/y\right)^3+\left(x/y\right)^2+\left(x/y\right)+1=0.$ It means that if $\chi$ is a integer such that $\chi^3+\chi^2+\chi+1=0$, that exists by assumption, then $b=\chi a$, $c=\chi^2 a$ and $d=\chi^3 a$. It's enough to say that $ a+b+c+d=a(\chi^3+\chi^2+\chi+1)=0$ and \[a^{2013}+b^{2013}+c^{2013}+d^{2013}=a^{2013}(\chi^{3\cdot 2013}+\chi^{2\cdot 2013}+\chi^{2013}+1)=0.\]Let's go back to the divisibility in $\mathbb{Z}$: clearly $0<a+b+c+d<4p$, implying that $a+b+c+d \in \{p,2p,3p\}$. But in every case $x \equiv x^{2013}\pmod{k}$ for every integer $x$ and $k \in \{1,2,3\}$: indeed $p$ has to be at least $5$ so that $\text{gcd}(k,p)=1$ and the conclusion follows by Chinese remainder theorem.