I will use "$N$" for the NAND function.$$ab+a'b' = ((a'+b')(a+b))'=(a'+b')N(a+b)$$ Now, $$a'+b'=(ab)'=aNb$$ and $$a+b=(a'b')'=(a')N(b')$$ and finally $$x'=(xx)'=xNx$$ so we get $$ab+a'b'=\Big(aNb\Big)N\Big((aNa)N(bNb)\Big)$$
I will use "$N$" for the NAND function.$$ab+a'b' = ((a'+b')(a+b))'=(a'+b')N(a+b)$$ Now, $$a'+b'=(ab)'=aNb$$ and $$a+b=(a'b')'=(a')N(b')$$ and finally $$x'=(xx)'=xNx$$ so we get $$ab+a'b'=\Big(aNb\Big)N\Big((aNa)N(bNb)\Big)$$