= Solution
Let $Z=Z(G)$. The inclusion $\mathbb F_p[[Z]]\subseteq Z(\mathbb F_p[[G]])$ is immediate. For the reverse inclusion, project a central element to every finite group algebra $\mathbb F_p[G/U]$, where $U$ ranges over open normal subgroups. Its coefficients are constant on conjugacy classes. Compatibility as $U$ shrinks shows that a nonzero coefficient can persist only on an element with finite conjugacy class in $G$: an infinite conjugacy orbit splits into arbitrarily large p-power collections in finer quotients, whose fibre sums vanish in characteristic $p$.
In a p-valued group, an element with finite conjugacy class is central. Indeed, its centralizer is open, so some p-power of every element centralizes it; the p-valuation and the leading commutator identity then force the original commutators to vanish. Thus the finite-conjugacy center is $Z$, and the compatible finite-quotient expansions are supported on $Z$. This proves the <Center of an Iwasawa algebra of a complete p-valued group>:
$$
\boxed{Z\bigl(\mathbb F_p[[G]]\bigr)=\mathbb F_p[[Z(G)]].}
$$
For $G=1+pM_2(\mathbb Z_p)$, a matrix commuting with every elementary principal-congruence matrix commutes with the full matrix algebra and is scalar. Hence
$$
Z(G)=\{uI_2:u\in1+p\mathbb Z_p\}\cong\mathbb Z_p.
$$
Consequently
$$
\boxed{Z\bigl(\mathbb F_p[[G]]\bigr)
\cong\mathbb F_p[[\mathbb Z_p]]
\cong\mathbb F_p[[T]].}
$$
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