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$(p\rightarrow q)\vee(p\rightarrow r)\equiv(\sim p\vee q)\vee(\sim p\vee r)$

Now we can change propositional operator into boolean operator for easy calculation

$\wedge\equiv\cdot,\vee\equiv +$

$(p\rightarrow q)\vee(p\rightarrow r)\equiv(\bar {p}+ q)+(\bar{p}+ r)$

$(p\rightarrow q)\vee(p\rightarrow r)\equiv(\bar {p}+ q)+(\bar{p}+ r)$

$(p\rightarrow q)\vee(p\rightarrow r)\equiv\bar {p}+ (q+ r)$

we can write like this

$(p\rightarrow q)\vee(p\rightarrow r)\equiv\sim {p}\vee (q\vee r)$

Hence

${\color{Magenta}{(p\rightarrow q)\vee(p\rightarrow r)\equiv p\rightarrow(q\vee r)} }$

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