On complete problems for $P/poly$ and $BPP$
There are no $P/Poly$ complete languages (relative to polynomial time or even computable reductions).

Suppose for the purpose of contradiction that there exists some language $L$ which is complete for $P/Poly$. In that case the mapping $f:P/Poly\rightarrow \mathbb{N}$, which maps any language in $P/Poly$ to its corresponding reduction to $L$ (we refer to the encoding of this reduction as a natural number), is an injection from $P/Poly$ to the natural numbers. This of course leads to a contradiction, since $P/Poly$ is of cardinality $2^{\aleph_0}$.

In light of Ricky Demer's comment, we should note that since computable reductions are of no use to $P/Poly$, one needs to introduce the notion of non-uniform reductions. One example is projection reductions.

A family of boolean functions $f_n(x_1,...,x_n)$ is said to be projection reducible to the family $g_n(y_1,...,y_n)$ if for all $n\in\mathbb{N}$ there exists some $m\le p(n)$ with polynomial $p$, and a rewiring of $g$'s inputs, $\sigma:\left\{y_1,...,y_m\right\}\rightarrow\left\{0,1,x_1,\overline{x_1},...,x_n,{\overline{x_n}}\right\}$, such that $f_n(x_1,...,x_n)=g_m\left(\sigma(y_1),...,\sigma(y_m)\right)$. Intuitively, this means that given a circuit for $g$ (perhaps with polynomialy more inputs), we can compute $f$ just be rewiring the inputs accordingly. We denote this by $f\le_{proj} g$. 

Note that if $g\in P/Poly$ and $f\le_{proj} g$ then $f\in P/Poly$, the advice for input $x$ would be the rewiring $\sigma$, together with the $m(|x|)$ inputs circuit for $g$. 

The following is a (very) short sketch of the proof that the circuit value problem, $CVP$, which takes as input a circuit $\mathcal{C}$ and input $x$, and asks whether $\mathcal{C}(x)=1$ is $P/Poly$ hard under such reductions. Given $L\in P/Poly$, let $\mathcal{C}_n$ be the polynomial circuit family for $L$. For any string $x$, You can hardwire $\mathcal{C}_{|x|}$ as input to $CVP$ (using the constants $0/1$), and use the rest of the variables to input $x$.