Solving all option one by one we get following truth table,
(1)
| p=r | q | p | r∨( p)</td><tdstyle="text−align:center;">(p∧q)</td><tdstyle="text−align:center;">(p∧q)∨r | r∨( p)⇒(p∧q)∨r |
| T | F | F | T | F | T | T |
| F | T | T | T | F | F | F |
| T | T | F | T | T | T | T |
| F | F | T | T | F | F | T |
(2)</p><tableborder="1"cellpadding="1"cellspacing="1"style="width:500px;"><tbody><tr><tdstyle="text−align:center;">p</td><tdstyle="text−align:center;">~ p</td><tdstyle="text−align:center;">r ∨ \left(~ p\right)</td><tdstyle="text−align:center;">q = r</td><tdstyle="text−align:center;">(p∧q)
\left(p ∧ q\right)∨r | r∨( p) ⇒ \left(p ∧ q\right)∨r | | T | F | T | T | T | T | T |
| F | T | T | T | F | T | T |
| T | F | F | F | F | F | T |
| F | T | T | F | F | F | F |
(3)</p><tableborder="1"cellpadding="1"cellspacing="1"style="width:500px;"><tbody><tr><tdstyle="text−align:center;">p</td><tdstyle="text−align:center;">q</td><tdstyle="text−align:center;">r = ~ p</td><tdstyle="text−align:center;">r ∨ \left(~ p\right)
\left(p ∧ q\right) | \left(p ∧ q\right) ∨ r</td><tdstyle="text−align:center;">r ∨ \left(~ p\right) ⇒ \left(p ∧ q\right) ∨ r</td></tr><tr><tdstyle="text−align:center;">T</td><tdstyle="text−align:center;">T</td><tdstyle="text−align:center;">F</td><tdstyle="text−align:center;">F</td><tdstyle="text−align:center;">T</td><tdstyle="text−align:center;">T</td><tdstyle="text−align:center;">T</td></tr><tr><tdstyle="text−align:center;">F</td><tdstyle="text−align:center;">T</td><tdstyle="text−align:center;">T</td><tdstyle="text−align:center;">T</td><tdstyle="text−align:center;">F</td><tdstyle="text−align:center;">T</td><tdstyle="text−align:center;">T</td></tr><tr><tdstyle="text−align:center;">T</td><tdstyle="text−align:center;">F</td><tdstyle="text−align:center;">F</td><tdstyle="text−align:center;">F</td><tdstyle="text−align:center;">F</td><tdstyle="text−align:center;">F</td><tdstyle="text−align:center;">T</td></tr><tr><tdstyle="text−align:center;">F</td><tdstyle="text−align:center;">F</td><tdstyle="text−align:center;">T</td><tdstyle="text−align:center;">T</td><tdstyle="text−align:center;">F</td><tdstyle="text−align:center;">T</td><tdstyle="text−align:center;">T</td></tr></tbody></table><p>(4) | p | p | q | r∨( p) | r= q | \left(p ∧ q\right)</td><tdstyle="text−align:center;">(p∧q)∨r | r∨( p)⇒(p∧q)∨r | | F | T | T | F | F | T | T | T | | F | T | F | T | T | F | T | T | | T | F | T | T | F | F | F | F | | T | F | F | T | T | F | T | T | Now final answer is option C.
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