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Solution

Check option (a):

$(p\to r)\land(q\to r)$

$=(\neg p\lor r)\land(\neg q\lor r)$

$=(\neg p\land\neg q)\lor r$

$=\neg(p\lor q)\lor r$

$=(p\lor q)\to r$

So, option (a) is logically equivalent.

Option (b) is also equivalent because if $p$ and $q$ have same truth value, then $\neg p$ and $\neg q$ also have same truth value.

Option (c) is the standard form of biconditional:

$p\leftrightarrow q\equiv(p\land q)\lor(\neg p\land\neg q)$

So, option (c) is logically equivalent.

Now check option (d):

$((p\land q)\to r)=\neg(p\land q)\lor r=\neg p\lor\neg q\lor r$

But,

$(p\to r)\land(q\to r)=(\neg p\lor r)\land(\neg q\lor r)$

Both are not always same.