You can rewrite c∧(a→b) as c∧((a)∨b). Use de Morgan's law to find that
(1) c∧((a)∨b) = c∧((a∧(b))) = c∧(a|(b)).
Now, observe that since
(2) d = d|d,
can be expressed in terms of the NAND operator. Therefore, ∧ can also be expressed in terms of the NAND operator since
(3) e∧f = (e|f).
Substituting the identities (2) and (3) into (1) as required will give an expression for c∧(a→b) which uses only NAND.