i am wondering if there is a complete solution for the equation $a^2+pb^22c^22kcd+(p+k^2)d^2=0$ in which $a,b,c,d,k$ are integer(not all zero) and $p$ is odd prime.

I'm not getting much 2adic information for this one, but it should be easy enough to check all solutions mod 8 and mod 16 and see what happens. To restrict anything, one property requires $p \equiv \pm 3 \pmod 8$ and the other requires $p \equiv 3 \pmod 4.$ Put them together, when $$p \equiv 3 \pmod 8 $$ and $$ p  k, $$ then all four of your letters $$ a,b,c,d = 0.$$ The proof uses two flavors of anisotropy for binaries. The assumption is that at least one of $ a,b,c,d $ is nonzero and $\gcd(a,b,c,d) = 1.$ First we have forced $a^2  2 c^2 \equiv 0 \pmod p,$ so $a,c \equiv 0 \pmod p$ as $(2  p) = 1.$ But then $ p b^2 + p d^2 \equiv 0 \pmod {p^2},$ or $ b^2 + d^2 \equiv 0 \pmod p,$ so $b,d \equiv 0 \pmod p$ as $(1  p) = 1.$ So $ p  \gcd(a,b,c,d)$ contrary to assumption. Otherwise, given a fixed $(p,k)$ once you have a nontrivial solution you get infinitely many using automorphs of the indefinite part in variables $(c,d).$ That is, there may be many parametrized families of solutions of one type or another. But you can figure some of those out with a computer algebra system more easily than I can by hand. The next interesting case is when $12 k^2 + 8p$ is a square, which means that the binary form $T(c,d)=2c^2+2kcd(p+k^2)d^2$ factors. So $3 k^2 + 2p$ is a square, which is not possible for even $k,$ so $k$ is odd and $2p \equiv 6 \pmod 8,$ or $p \equiv 1 \pmod 4.$ Unless $p=3$ we also need $p \equiv 1 \pmod 3,$ or $p \equiv 1 \equiv 11 \pmod {12}.$ For example, with $p=11, k=1,3 k^2 + 2p = 25, p + k^2 = 12,$ we have $$ a^2+11b^22c^22cd+12d^2 = a^2+11b^22(c2d)(c+3d).$$ The value of the factorization is that we can take, for instance, $c = 2 d + 1, c + 3 d = 5 d + 1,$ and $$ a^2+11b^22(5d+1) = 0.$$ Now $a^2 + 11 b^2$ is not even unless it is also divisible by $4.$ We also need $ a^2 \equiv b^2 \equiv 1 \pmod 5.$ Put them together, we have a parametrized solution of sorts, with $$ a \equiv 1,4 \pmod 5, \; \; b \equiv 1,4 \pmod 5, \; \; a \equiv b \pmod 2$$ take $c = 2 d + 1$ and $$ d = \frac{ a^2+11b^22}{10}.$$ 


Perhaps it would help if we knew where the question is coming from. For what it's worth, you can write the equation in the form $$ a^2 + (ckd)^2 + pb^2 + pd^2 = 3c^2, $$ so you are looking at a parametrized subset of the equation $$ A^2 + B^2 + pC^2 + pD^2 = 3c^2. $$ If $p \equiv 1 \bmod 4$, then $p(B^2+D^2) = R^2 + S^2$ is a sum of two squares, and your solutions must occur among those of $$ R^2 + S^2 + T^2 + U^2 = 3c^2. $$ Both quadrics can be parametrized by the standard method of sweeping lines if you know one solution. For arbitrary primes $p$ such a solution seems to be difficult to find. And even armed with such a parametrization you then would have to figure out which of them satisfy the additional conditions coming from the original equation. 


There are lots! Here is a selection: $(a,b,c,d,k,p)=(3,3,4,1,1,3)$ $(a,b,c,d,k,p)=(3,3,4,1,7,3)$ $(a,b,c,d,k,p)=(3,3,4,2,2,3)$ 


For the equation: $$a^2+pb^2+(p+k^2)z^2=2c^2+2kcz$$ If the number $k$ is the problem any, and $p$ is such as this: $p=\frac{t^2}{2}1$ Then the solution can be written: $$a=\pm{t}n^2+2(tpr\mp(p+1)kj)ns(2p(p+1)kjr\pm{t}((p+1)(p+k^2)j^2+pr^2))s^2$$ $$b=\pm{t}n^22(tr\pm(p+1)kj)ns+(2(p+1)kjr\mp{t}((p+1)(p+k^2)j^2+pr^2))s^2$$ $$z=2(p+1)j((p+1)kjstn)s$$ $$c=(p+1)(n^2+((p+1)(p+k^2)j^2+pr^2)s^2)$$ $n,s,j,r$  integers which we are set. If you can represent numbers as: $p=3k^2t^2$ This decision when the coefficients are related through the equation of Pell. $t^23k^2=p$ To simplify calculations we will make this change. $$x=(\pm{t}2k)n^2+2j(t\mp3k)ns(2kj^2+2kpe^2\pm{t}(pe^22j^2))s^2$$ $$y=(\pm{t}2k)n^2+2j(2t\mp3k)ns(8kj^2+2kpe^2\pm{t}(pe^22j^2))s^2$$ $$r=2e(tn3kjs)s$$ $$f=n^2+(pe^22j^2)s^2$$ Then the solution can be written: $$a=pr^2+(p+k^2)f^2xy$$ $$b=r(x+y)$$ $$z=f(x+y)$$ $$c=pr^2+(p+k^2)f^2+x^2$$ $n,s,e,j$  integers which we ask. 

