How to pass multiple arguments in processStartInfo?

System.Diagnostics.Process process = new System.Diagnostics.Process();
System.Diagnostics.ProcessStartInfo startInfo = new System.Diagnostics.ProcessStartInfo();
startInfo.WindowStyle = System.Diagnostics.ProcessWindowStyle.Normal;
startInfo.FileName = "cmd.exe";
startInfo.Arguments = @"/c -sk server -sky exchange -pe -n CN=localhost -ir LocalMachine -is Root -ic MyCA.cer -sr LocalMachine -ss My MyAdHocTestCert.cer"

use /c as a cmd argument to close cmd.exe once its finish processing your commands

startInfo.Arguments = "/c \"netsh http add sslcert ipport=127.0.0.1:8085 certhash=0000000000003ed9cd0c315bbb6dc1c08da5e6 appid={00112233-4455-6677-8899-AABBCCDDEEFF} clientcertnegotiation=enable\"";

and...

startInfo.Arguments = "/c \"makecert -sk server -sky exchange -pe -n CN=localhost -ir LocalMachine -is Root -ic MyCA.cer -sr LocalMachine -ss My MyAdHocTestCert.cer\"";

The /c tells cmd to quit once the command has completed. Everything after /c is the command you want to run (within cmd), including all of the arguments.

It is purely a string:

startInfo.Arguments = "-sk server -sky exchange -pe -n CN=localhost -ir LocalMachine -is Root -ic MyCA.cer -sr LocalMachine -ss My MyAdHocTestCert.cer"

Of course, when arguments contain whitespaces you'll have to escape them using \" \", like:

"... -ss \"My MyAdHocTestCert.cer\""

See MSDN for this.