Explanation of a particular piece of C*-algebra notation For this question, $A$ is a given (possibly non-unital) $C^*$-algebra and $p$ is an element of the multiplier algebra, $M(A)$, satisfying $p^2=p=p^*$ (ie is a projection). $M(A)$ is most easily described as linear right-$A$-module maps $A\to A$ with an adjoint with respect to the $A$-valued inner product $\langle a , b \rangle_A := a^*b$ on $A$.
Recall that $A$ is an ideal in $M(A)$, so elements of $A$ can be considered as multipliers (acting on the left: $m_a \cdot b = ab$). If $p\in A$, then one can consider the corner $pAp$, which is the subalgebra of elements of the form $pap$, for $a$ in $A$. Likewise, there is the two-sided ideal $ApA$, which is the span of elements of the form $a_1pa_2$, for $a_1,a_2 \in A$. There is also the right ideal $pA$, with elements $pa$.
However, these notations are also used for arbitrary projections $p$ in $M(A)$. It is not entirely clear to me what sets $pAp$ and $ApA$ are supposed to be. The right ideal $pA$ is not so mysterious, since one can describe it as the image of the linear map $p\colon A \to A$ (and, in that case, $A \simeq pA \oplus (id - p)A$ as  C^*-algebras topological vector spaces).

What is a concrete description of $pAp$ and $ApA$ when $p$ is some projection in $M(A)$?

 A: You say that $M(A)$ is "most easily" described, and I have to disagree. In my opinion, $M(A)$ is most easily described by considering any faithful non-degenerate representation $A\subset B(H)$, and taking $$\tag1M(A)=\{T\in B(H):\ Ta\in A,\ aT\in A,\ \text{ for all } a\in A\}.$$
Since $A\subset M(A)$ as an ideal, the products $pap$ and $apb$ are perfectly well defined for any $p\in M(A)$ and $a,b\in A$. I'm not entirely sure what you mean by "concrete": most elements in most C$^*$-algebras are not "concrete" if you mean that you can write them explicitly (note that this assertion is even true in the case of the real numbers). 
One can play the game of using a concrete construction of $M(A)$ and try to write $pAp$ "explicitly"; this could be an interesting exercise in formalism, but it won't clarify anything. For instance, using your picture you first need to see $A$ inside $M(A)$. As you said, $a$ corresponds to $m_a$. So if $p\in M(A)$ is a projection, you have $p^2=p^*=p$. The fact that $p=p^*$ means that $p(a)^*b=a^*p(b)$ for all $a,b\in A$. And $pap:A\to A$ is the map 
$$
(pap)b=p(ap(b))=p(a)p(b)
$$
(remember that $p$ is a right $A$-module map). Not very concrete. 
Finally, you say that $A\simeq pA\oplus (I-P)A$ "as C$^*$-algebras"). Note that $pA$ is a right ideal, not a subalgebra. 
