Comultiplication of elements of partition of unity Let $F(G)$ be the algebra of functions on a finite quantum group $G$ (so that $F(G)$ is a finite dimensional $\mathrm{C}^*$-Hopf algebra).
Suppose that $\{p_i:i=0,\dots,d-1\}\subset F(G)$ is a partition of unity, in other words $p^2_i=p_i^*=p_i$,
$$p_ip_j=p_jp_i=\delta_{i,j}\,p_i,$$
 and
$$\sum_{i=0}^{d-1}p_i=\mathbf{1}_G:=1_{F(G)},$$
the unit of $F(G)$. 
Edit: The following condition was added after Konstantinos' answer:

Suppose that a state $\nu\in M_p(G):=\mathcal{S}(F(G))$ has the
  property that for all projections $q\in F(G)$, there exists $k_q$ such
  that $\nu^{\star k_q}(q)\neq 0$, where $$\nu\star \nu=(\nu\otimes
 \nu)\circ \Delta.$$

Suppose furthermore that $\nu\in M_p(G)$ has the property that:
$$\nu(p_i)=\begin{cases}1 & \text{ if }i=1\\ 0 & \text{else}\end{cases},$$
and we also have that, where $\varepsilon\in M_p(G)$ is the counit:
$$\varepsilon(p_i)=\begin{cases}1 & \text{ if }i=0\\ 0 & \text{else}\end{cases}.$$
Furthermore,
$$(\nu\otimes I_{F(G)})\circ \Delta(p_i)=:T_\nu(p_i)=p_{i-1},$$
with $T_\nu(p_0)=p_{d-1}$.
Note that $\Delta$ is a *-homomorphism, and, where $\int_G:=h\in M_p(G)$ is the Haar state of $F(G)$, we can show that:
$$\int_Gp_i=\frac{1}{d}.$$
Is it the case that

$$\Delta(p_i)=\sum_{k=0}^{d-1}p_{i-k}\otimes p_k?$$

If $F(G)$ is commutative, this condition holds.
 A: This answer was given before the edit
(with the understanding that under the stated assumptions the comultiplication described in the OP is cocommutative for the $d$ idempotents $p_i$ and $k$ is an algebraically closed field of char zero). 


*

*An example where "it is the case":
Consider the finite group $N$ and the cyclic group $C_d$ of order $d$. Then $k(N\times C_d)=kN\otimes kC_d$ is the group hopf algebra of the group $N\times C_d$. If we take its dual let us set:
$$
F(G):=\big(k(N\times C_d)\big)^*\cong\big(kN\otimes kC_d\big)^*\cong(kC_d)^*\otimes(kN)^*\cong kC_d\otimes (kN)^*
$$
because $(kC_d)^*\cong kC_d$ as hopf algebras, for any finite abelian group. Inside $(kC_d)^*$ the multiplication and the comultiplication are exactly as in the OP.
(Actually, any finite abelian group $H$ of order $d$, in place of $C_d$ would do the job). 

*A counterexample (where "it is not the case"):
Consider the finite group $N$ and the finite non-abelian group $H$ of order $d$. Then $k(N\times H)=kN\otimes kH$ is the group hopf algebra of the group $N\times H$. If we take its dual let us set: 
$$
F(G):=\big(k(N\times H)\big)^*\cong\big(kN\otimes kH\big)^*\cong(kH)^*\otimes(kN)^*
$$
Inside $(kH)^*$ the multiplication is isomorphic to the one described in the OP (i.e. the orthogonal idempotents providing a partition of unity) but the comultiplication cannot be the one suggested in the OP. The reason is that since $kH$ is non-commutative then its dual hopf algebra $(kH)^*$ cannot be cocommutative. 
A: These are additions to Konstantinos' answer and was given before the edit
A Counterexample
This is inspired from here and here.
Let $G=S_3\times C_2$ and $\nu=\frac{1}{2}(\delta^{(e,1)}+\delta^{((12),1)})$.
Now define $S_0=\{(e,0),((12),0),((13),1),((23),1),((123),1),((132),1)\}$ and $S_1=G\backslash S_0$.
Consider $p_0=\mathbf{1}_{S_0}$ and $p_1=\mathbf{1}_{S_1}$.
These projections have all the properties given above.
Consider 
$$p_1=\delta_{((123),0)}+ \cdots.$$
Note
$$\begin{align*}
\Delta(p_1)&=\Delta(\delta_{((123),0)}+\cdots)
\\&=\underbrace{\delta_{((132),0)}\otimes \delta_{((132),0)}}_{\in p_1F(G)\otimes p_1F(G)}+\cdots,\end{align*}$$
and so 
$$\Delta(p_1)\neq p_0\otimes p_1+p_1\otimes p_0.$$
A Missing Condition
I am missing the following condition. We suppose in addition that for all projections $q\in F(G)$, there exists $k\in \mathbb{N}$ such that $\nu^{\star k}(q)\neq 0$, where 
$$\nu\star \nu=(\nu\otimes\nu)\circ \Delta.$$
This will be added to the question.
In the classical case, where $F(G)$ is commutative, it can be shown that $p_0=\mathbf{1}_N$, where $N\lhd G$, $p_1=\mathbb{1}_{Ng}$, and $p_m=\mathbb{1}_{Ng^m}$, i.e. the $p_i$ are indicator functions on cosets of the normal subgroup $N\lhd G$. Furthermore, $G/N\cong C_d$, and so with the missing condition, in the classical case, the projections satisfy:
$$\Delta(p_i)=\sum_{k=0}^{d-1}p_{i-k}\otimes p_k.$$
I will edit the original question.
